Answers
Given data:
* The mass of the car is m = 1200 kg.
* The distance up to which Tyler can see is d = 15 m.
* The coefficient of friction between the road and the car is,
\(\mu=0.67\)* The force applied by Tyler on the car is,
\(F=1.3\times10^4\text{ N}\)Solution:
The diagrammatic representation of the given case is,
(a). The weight of the car is,
\(W_{fg}=mg\)where g is the acceleration due to gravity,
Substituting the known values,
\(\begin{gathered} W_{fg}=1200\times9.8 \\ W_{fg}=11760\text{ N} \end{gathered}\)Thus, the weight of the car is 11760 N.
(b). The normal force acting on the car is,
\(\begin{gathered} N=W_{fg} \\ N=11760\text{ N} \end{gathered}\)The frictional force acting on the car is,
\(\begin{gathered} F_r=\mu N \\ F_r=0.67\times11760 \\ F_r=7879.2\text{ N} \end{gathered}\)Thus, the net force acting on the car is,
\(\begin{gathered} W_{\text{net}}=F-F_r \\ W_{\text{net}}=1.3\times10^4-7879.2 \\ W_{\text{net}}=13000-7879.2 \\ W_{\text{net}}=5120.8\text{ N} \end{gathered}\)Thus, the net force acting on the car is 5120.8 N.
(c). According to Newton's second law, the acceleration of the car is,
\(\begin{gathered} W_{\text{net}}=ma \\ a=\frac{W_{\text{net}}}{m} \end{gathered}\)Substituting the known values,
\(\begin{gathered} a=\frac{5120.8}{1200} \\ a=4.27ms^{-2} \end{gathered}\)Thus, the acceleration of the car is 4.27 meters per second squared.
(d). The initial velocity of the car is zero, by the kinematics equation, the time taken by the car to reach the gas station is,
\(d=ut+\frac{1}{2}at^2\)where u is the initial velocity,
Substituting the known values,
\(\begin{gathered} 15=0+\frac{1}{2}\times4.27\times t^2 \\ 15=2.14t^2 \\ t^2=\frac{15}{2.14} \\ t^2=7 \end{gathered}\)By taking the square root,
\(t=2.6\text{ s}\)Thus, the time taken by Tyler to reach the gas station is 2.6 seconds.
Related Questions
Object 1 with mass 1=3.25 kg
is held in place on an inclined plane that makes an angle
of 40.0∘
with the horizontal. The coefficient of kinetic friction between the plane and the object is 0.535.
Object 2 with mass 2=4.75 kg
is connected to object 1 with a massless string over a massless, frictionless pulley. The objects are then released.
Calculate the magnitude
of the initial acceleration.
Calculate the magnitude
of the tension in the string once the objects are released.
Answers
The magnitude of the initial acceleration of the object is 4.2 m/s².
The tension in the string once the object starts moving is 13.65 N.
What is the magnitude of the initial acceleration?The magnitude of the initial acceleration of the object is calculated by applying Newton's second law of motion as follows;
F(net) = ma
m₂g - μm₁g cosθ = a(m₁ + m₂)
where;
m₁ and m₂ are the masses of the blocksg is acceleration due to gravityμ is coefficient of frictionθ is the angle of inclinationa is the acceleration(4.75 x 9.8) - (0.535 x 3.25 x 9.8 x cos40) = a(3.25 + 4.75)
33.5 = 8a
a = 33.5/8
a = 4.2 m/s²
The tension in the string once the object starts moving is calculated as;
T = m₁a
T = 3.25 x 4.2
T = 13.65 N
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2 A rectangular storage tank 4 m long by 3 m wide is filled with paraffin to a depth
of 2 m. Calculate:
a the volume of paraffin
c the weight of paraffin
b the mass of paraffin
d the pressure at the bottom of the tank due
to the paraffin
1m
Answers
For a rectangular storage tank filled with paraffin to a depth of 2 m, the volume, weight, mass of paraffin, and pressure at the bottom of the tank are:
a. The volume is 24 m³.
b. weight is 240,000 N,
c. mass is 24,490 kg, and
d. pressure is 23,530 Pa.
a) The volume of paraffin in the rectangular storage tank can be calculated using the formula:
Volume = Length x Width x Depth
Given:
Length = 4 m
Width = 3 m
Depth = 2 m
Substituting the values into the formula, we have:
Volume = 4 m x 3 m x 2 m
Volume = 24 m³
Therefore, the volume of paraffin in the tank is 24 cubic meters.
b) The weight of the paraffin can be calculated using the formula:
Weight = Volume x Density x Acceleration due to gravity
The density of paraffin varies, but we can assume a typical value of 10,000 kg/m³. The acceleration due to gravity is approximately 9.8 m/s². Substituting these values into the formula:
Weight = 24 m³ x 10,000 kg/m³ x 9.8 m/s²
Weight = 240,000 N
Therefore, the weight of the paraffin in the tank is 240,000 Newtons.
c) The mass of the paraffin can be calculated using the formula:
Mass = Density x Volume
Substituting the given values:
Mass = 10,000 kg/m³ x 24 m³
Mass = 24,490 kg
Therefore, the mass of the paraffin in the tank is 24,490 kilograms.
d) The pressure at the bottom of the tank due to the paraffin can be calculated using the formula:
Pressure = Weight / Area
The area of the bottom of the tank is equal to the length multiplied by the width. Substituting the values:
Area = 4 m x 3 m
Area = 12 m²
Pressure = 240,000 N / 12 m²
Pressure = 20,000 Pa
Therefore, the pressure at the bottom of the tank due to the paraffin is 20,000 Pascals (Pa).
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A sharpshooter fires a rifle while standing with the butt of the gun against her shoulder. If the forward momentum of a bullet is the same as the backward momentum of the gun, why isn't it as dangerous to be hit by the gun as by the bullet? (Select all that apply.)
The butt distributes the recoil force over an area much larger than that of the bullet.
The rifle has a much lower speed than the bullet.
The rifle has a much smaller kinetic energy than the bullet.
The rifle has much less mass than the bullet.
The rifle has much more total momentum than the bullet.
Answers
It as dangerous to be hit by the gun as by the bullet because of the following;
(A) The butt distributes the recoil force over an area much larger than that of the bullet.
(B) The rifle has a much lower speed than the bullet.
What is principle of conservation of linear momentum?
The principle of conservation of linear momentum states that in an isolated system, the total momentum of the system is conserved.
That is the sum of the initial momentum is equal to the sum of the final momentum.
momentum of the gun = momentum of the bullet
Mu = mU
where;
M is mass of the gunu is the velocity of the gunm is mass of the bulletU is the velocity of the bulletIf the forward momentum of a bullet is the same as the backward momentum of the gun, the speed of the gun will be smaller than the speed of the bullet since the mass of the gun is bigger than mass of the bullet.
We cannot conclude on the kinetic energy, since it depends on both mass and velocity.
Finally, the butt distributes the recoil force over an area much larger than that of the bullet, since the butt has a larger surface area and will hit more surface area than the bullet.
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URGENT PLEASE... SAVE ME FROM THIS HW
An object at rest starts accelerating. If it travels 40 meters to end up going 20 m/s, what was its acceleration.
Answers
^ 3.0
Explanation: The acceleration is the change in
velocity per unit time.
Here the initial velocity is O the final velocity is
20.
So this formula is applicable.
V^2= u^2 + 2as
20^2 = 0^2+ 2a (s)
Answer:
The car's acceleration is 5m/s^2
Explanation:
acceleration is the change in velocity per unit time.
so this formula will work.
V^2= u^2 + 2as
20^2 = 0^2+ 2a (s)
Determine the energy released per kilogram of fuel used.
Given MeV per reaction, calculate energy in joules per kilogram of reactants.
Consider 1 mile of tritium plus 1 mole of deuterium to be a mole of "reactions" ( total molar mass = 5 grams)
Answers
The energy released per kilogram of fuel used is 3.39 * 1014 J/Kg
Why is the energy released in a reaction?Energy is released in a reaction because of the breaking of bonds are well as formation of bonds.
The quantity of energy released in reactions differs according to the reaction type involved.
The energy released in nuclear reactions are far larger than that released in chemical reactions due to the release of nuclear energy from the nucleus.
The energy, E released in nuclear reactions is given by the formula below:
Energy per kilogram of reactants (in joules) = Energy per mole of reactants (in joules) / Total molar mass of reactants (in kg)
Energy per kilogram of reactants (in joules) = \(1.60218 x 10^-^1^3 joules\) / 5 grams
Energy per kilogram of reactants (in joules) = 3.39 * 1014 J/Kg
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The energy released per kilogram of fuel used is 5.632 × 10^-14 J/kg.
5 grammes, 0.005 kilogrammes, of reactants are provided.
Since 1 mile of tritium and 1 mile of deuterium are equal to 1 mole of "reactions," the sum of the "reactions" in 0.005 kilogrammes of reactants may be computed as follows:
Total moles of "reactions" in 0.005 kg of reactants = (0.005 kg / 5 g/mol)
= 0.001 mole.
The MeV per reaction must now be multiplied by the total number of "reactions" in order to get the total energy released by the "reactions."
Next, the energy must be converted from MeV to Joules.
MeV to Joules conversion factor is 1.6 10-13 J/MeV.
Total energy released = (MeV per reaction) x (number of reactions) x (conversion factor)
Total energy released = (17.6 MeV/reaction) x (0.001 mole) x (1.6 × 10^-13 J/MeV)
Total energy released = 2.816 × 10^-16 J
The total energy released by the "reactions" is 2.816 × 10^-16 J.
To determine the energy released per kilogram of fuel used, we need to divide the total energy by the mass of fuel used.
Total energy released per kilogram of fuel used = (total energy released) / (mass of fuel used)
Total energy released per kilogram of fuel used = (2.816 × 10^-16 J) / (0.005 kg)
Total energy released per kilogram of fuel used = 5.632 × 10^-14 J/kg
Therefore, the energy released per kilogram of fuel used is 5.632 × 10^-14 J/kg.
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A freight train car has a mass of 2,000 kilograms and an acceleration of 1.8 m/s/s. What is the average force behind that train car?
Answers
Answer:
F = 3600 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of force must be equal to the product of mass by acceleration.
ΣF = m*a
where:
F = force [N]
m = mass = 2000 [kg]
a = acceleration = 1.8 [m/s^2]
Now replacing:
F = 2000*1.8
F = 3600 [N]
An organ pipe closed at one end and open at the other has a length of 0.6 m.
(a) What is the longest possible wavelength for the interfering sound waves that can form a standing wave in this pipe?
(b) What is the frequency associated with this standing wave if the speed of sound
Answers
(a) The longest possible wavelength for the interfering sound waves that can form a standing wave is 2.4 m
b. , the frequency associated with the standing wave in this pipe is 142.9 Hz
How do we calculate?a.) The longest possible wavelength for the interfering sound waves that can form a standing wave in a pipe closed at one end and open at the other is four times the length of the pipe, or 4L.
L = 0.6 m, so the longest possible wavelength is:
λ_max = 4L = 4 * 0.6 m = 2.4 m
(b)
the frequency associated with the standing wave in this pipe is calculated as
v = fλ
v = 343 m/s
λ = 2.4 m
f = v/λ = 343 m/s / 2.4 m = 142.9 Hz
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What happens when data stored digitally are copied?
OA. Each copy made has worse quality than the original data did.
B. Data are copied quickly, easily, and perfectly.
C. Only one value of the digital data is copied perfectly.
D. Data are copied quickly, but they decrease in quality.
Answers
When data stored digitally are copied then data are copied quickly, easily, and perfectly.
Data stored digitally are quickly, easily and perfectly copied.
There is no effect on the quality of the data.It just copies the data exactly as the source and it is same with any other file type. The quality would only decrease if the file is transcoded to a different format or to the same format with a different data stream.Data is a collection of quantities, photos or videos, signs or symbols, on which operations are performed by a computer, which can be stored and transmitted in the form of electrical signals and recorded on magnetic, optical or mechanical recording media.
When you copy (or transfer) a file of any type - JPG or TIFF or RTF or DOC or PSD or whatever, you won't lose any quality.Since no quality of the file is lost, copying a data is an easy, quick and a perfect process.
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The 10/90 principle can help you take control of your situation in taking responsibility of what you can change rather than in being victim of what you cannot change. Give an example of a situation that can change for you in applying this principle.
Answers
The 10/90 principle can be a powerful tool for taking control of your situation and improving your life. By taking responsibility for what you can change and focusing on your reaction to the situation, you can make positive changes in your life and become the master of your own destiny.
The 10/90 principle refers to the idea that life is made up of 10% of what happens to you and 90% of how you respond to it. In other words, you may not be able to control what happens to you, but you can control your reaction to it. By taking responsibility for what you can change rather than being a victim of what you cannot change, you can take control of your situation and improve your life.One example of a situation where the 10/90 principle could be applied is losing a job. Losing a job can be a devastating experience, and it can be easy to feel like a victim in this situation. However, by applying the 10/90 principle, you can take control of your situation and make positive changes in your life.The first step in applying the 10/90 principle in this situation would be to take responsibility for what you can change. This could mean updating your resume, networking with others in your field, and applying for new jobs. By taking action and doing what you can to find a new job, you are taking control of your situation and improving your chances of finding a new job.
The second step would be to focus on your reaction to the situation. Instead of dwelling on the negative aspects of losing your job, try to focus on the positive aspects. This could mean using the extra time to pursue a new hobby or spend more time with family and friends. By focusing on the positive aspects of the situation, you are taking control of your reaction and improving your overall well-being.
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1. Explain home a dial test indicator (DTI) reading is adjusted when the gange is not positioned at right angle to the contact surface.
Answers
When using a dial test indicator (DTI), it is essential to ensure that the gauge is positioned at a right angle to the contact surface for accurate readings.
However, in certain situations, it may be challenging to achieve a perfect right angle alignment. In such cases, adjustments can be made to compensate for the misalignment and obtain accurate measurements.To adjust the DTI reading when the gauge is not positioned at a right angle to the contact surface, the following steps can be taken:Determine the misalignment angle: Measure the angle at which the DTI is misaligned from the right angle position. This can be done using a protractor or by estimating the deviation visually.Calculate the correction factor: Based on the misalignment angle, calculate the correction factor using trigonometric functions such as sine or cosine. The correction factor accounts for the difference between the actual displacement and the displacement measured by the DTI.Apply the correction factor: Multiply the correction factor by the DTI reading to adjust the measurement. This compensates for the misalignment and provides a more accurate reading.It's important to note that adjusting the DTI reading can introduce some degree of error, especially if the misalignment is significant. Therefore, it is always preferable to position the gauge at a right angle to the contact surface whenever possible to obtain the most precise measurements.
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After an unfortunate accident occurred at a local warehouse, you were contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure.The horizontal steel beam had a mass of 88.50 kg
per meter of length, and the tension in the cable was =11650 N
. The crane was rated for a maximum load of 500 kg
. If =5.580 m
, =0.522 m
, =1.350 m
, and ℎ=2.070 m
, what was the magnitude of L
(the load on the crane) before the collapse? The acceleration due to gravity is =9.810 m/s2
Answers
The magnitude of the load L on the crane before the collapse is 3211.81 N
To determine the magnitude of the load on the crane (L), we will need to make use of the equilibrium conditions of the torque.
It is always an ideal process to list out all the parameters given as this will let you understand how you can determine the answer to the question from the given parameters.
From the given information;
The tension in the cable = 11650 NThe length (d) = 5.580 mThe mass of the horizontal steel beam (M) = 88.50 kg/m (d)= 88.50 kg/m × 5.580 m= 493.83 kgDistance (s) = 0.522 mx = 1.350 m and h = 2.070 mAcceleration due to gravity = 9.81 m/s²From the question;
the angle at which the crane is positioned can be determined by taking the tangent of the angle θ. i.e.
\(\mathbf{tan \ \theta = \dfrac{h}{d-s}}\)
\(\mathbf{\theta = tan^{-1} \Big ( \dfrac{h}{d-s} \Big )}\)
\(\mathbf{\theta = tan^{-1} \Big ( \dfrac{2.070 }{5.580 - 0.522} \Big )}\)
\(\mathbf{\theta =22.26^0}\)
Consider the equilibrium conditions of the torques with respect to the magnitude of the load at point P.
∴
\(\mathbf{Tsin \theta (d-s) - W_L (d-x) -(Mg) (\dfrac{d}{2}) = 0}\)
By making the magnitude of the load \(\mathbf{W_L}\) the subject of the formula, we have:
\(\mathbf{W_L = \dfrac{Tsin \theta (d-x) -(Mg) (\dfrac{d}{2})}{ (d-s) } }\)
\(\mathbf{W_L = \dfrac{(11650 )sin (22.26) (5.580-1.350) -(88.50\times 9.81) (\dfrac{5.580}{2})}{ (5.580-0.522) } }\)
\(\mathbf{W_L = 3211.81 \ N }\)
Therefore, we can conclude that the magnitude of the load is 3211.81 N
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5. True or False: The medium moves along with a wave.
Answers
Answer:
false. the medium never moves
Explanation:
During a relay race, runner A runs a certain distance due north and then hands off the baton to runner B, who runs for the same distance in a direction south of east. The two displacement vectors A and B can be added together to give a resultant vector R. Which drawing correctly shows the resultant vector?
Answers
Answer:
d) 4
Explanation:
The image attached shows the options.
The resultant vector is the resultant of two or more vectors. The resultant vector is gotten by adding the sum of the displacement of the vectors together (that is the sum of all the individual vectors).
From the question, since runner A runs north and runner B runs east the resultant vector would be the sum of the displacement of vectors A and B. Also, the direction of the vector would be the north east starting from the beginning of vector A to end of vector B. The correct option is d) 4
what change in the gravitation occurs if the distance between two bodies is doubled by keeping the mass constant?
Answers
Explanation:
Force of gravity = G m1 m2 / r2
now double 'r'
Force = G m1 m2 / (2r) ^2
Force = 1/4 G m1 m2 / r^2 <=====this is 1/4 of the original force of gravity
. When a large truck
hits a small car, the
forces are equal.
• However, the small
.
car experiences a
much greater
change in velocity
much more rapidly
than the big truck.
Which vehicle ends up
with more damage?
Answers
Answer:
Car
Explanation:
Based on Newton 2nd law, energy are conserves. Meaning that if the Force is equal, the car with lower mass must be travelling in a much greater acceleration.
F = m.a
where,
a = Δv/Δt
When talking about energy, there are 2 factor: mass and velocity.
The change of Kinetic energy experience by the car is
ΔEk = 1/2.m.Δv²
Eventhough the car has smaller mass, notice that the velocity will be squared. In this case the velocity is the a more dominant factor. It means that energy absorbed by the car is much larger.
A small car with mass 0.610 kg
travels at constant speed on the inside of a track that is a vertical circle with radius 5.00 m
If the normal force exerted by the track on the car when it is at the top of the track (point B
) is 6.00 N
, what is the normal force on the car when it is at the bottom of the track (point A
)?
Express your answer with the appropriate units.
Answers
The normal force on the car when it is at the bottom of the track is 8.84 N.
At the top of the track (point B), the normal force (N) acting on the car is equal to the weight of the car (mg) plus the centrifugal force (mv²/r) acting outwards:
N = mg + mv²/r
where m is the mass of the car, g is the acceleration due to gravity, v is the speed of the car, and r is the radius of the track.
Since the car is traveling at constant speed, its acceleration is zero, so the centrifugal force is balanced by the force of gravity, and we have:
N = mg
Substituting the given values, we get:
6.00 N = (0.610 kg)(9.81 m/s²)
Solving for g, we get:
g = 9.68 m/s²
At the bottom of the track (point A), the normal force acting on the car is equal to the weight of the car minus the centrifugal force acting downwards:
N = mg - mv²/r
Substituting the known values, we get:
N = (0.610 kg)(9.81 m/s²) - (0.610 kg)(v²/5.00 m)
Since the car is traveling at constant speed, we can use the fact that its kinetic energy is equal to its gravitational potential energy to solve for v:
mg(2r) = (1/2)mv²
where 2r is the total distance traveled by the car (i.e., the circumference of the circle), so 2r = 2πr = 31.4 m. Solving for v, we get:
v = sqrt(2gr)
Substituting the known values, we get:
v = sqrt(2(9.68 m/s²)(5.00 m)) = 6.19 m/s
Substituting this value of v into the equation for N, we get:
N = (0.610 kg)(9.81 m/s²) - (0.610 kg)((6.19 m/s)²/5.00 m) = 8.84 N
Therefore, the normal force on the car when it is at the bottom of the track is 8.84 N.
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At a typical bowling alley the distance from the line where the ball is released (foul line) to the first pin is 60ft. Assume it takes 5.0s for the ball to reach the pins after you release it, if it rolls without slipping and has a constant translational speed. Also assume the ball weights 12lb and has a diameter of 8.5 inches.
A- Calculate the rotation rate of the ball in rev/s
B- What is the total kinetic energy in pound-feet? Ignore the finger holes and treat the bowling ball as a uniform sphere
Answers
A) The rotation rate of the ball in rev/secs is ; 10.79 rev/secs
B) The total kinetic energy of in pound-feet ignoring the finger holes is : 37.6157 pound-feet
Given data :
mass of ball = 12 Ib
diameter of ball = 8.5 inches
Radius = 4.25 inches = 0.10795 m
Time = 5 secs
distance travelled = 60 ft = 18.288 m
A) Determine the rotation rate of the ballFirst step : calculate the velocity of ball
V = distance / time
= 18.288 / 5 = 3.66 m/s
Next step :
angular velocity ( w ) = V / r
= 3.66 / ( 0.10795 ) = 33.90 rad/sec
convert to rev/sec = 33.90 / π
= 10.79 rev/secs
B) Determine the total kinetic energy
given that ball rolls without slipping
Vcon = Rw
Total K.E = 1/2 IW² + 1/2 mv²
= 7/10 MR²W²
Total K.E = 7/10 * 5.44 * ( 0.10795 )² * 33.90²
= 50.99 ≈ 51 J
convert to pound-feet = 37.6157 pound-feet
Hence we can conclude that The rotation rate of the ball in rev/secs is ; 10.79 rev/secs and The total kinetic energy of in pound-feet ignoring the finger holes is : 37.6157 pound-feet.
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L 2.5.2 Test (CST): The Brain and the Body
Question 4 of 10
The
lobes are important for decision making and judgment.
A. frontal
B. temporal
O c. parietal
D. occipital
SUBMIT
Answers
Answer:
I believe it's A frontal. Hope this can help
The frontal lobes are important for decision-making and judgment. The correct answer is A.
What are frontal lobes?The frontal lobes are located at the front of the brain and are involved in a variety of cognitive functions, including reasoning, problem-solving, planning, decision-making, and judgment. The prefrontal cortex, which is located in the frontal lobes, is especially important for these functions.
In addition, the frontal lobes are involved in the control of voluntary movements, speech production, and the regulation of emotions and social behavior. Damage to the frontal lobes can lead to a variety of cognitive and behavioral deficits, including difficulty with decision-making and judgment, impaired impulse control, and changes in personality.
The temporal lobes are located on the sides of the brain and are involved in the processing of auditory information, as well as memory formation and recognition of faces and objects. The parietal lobes are located at the top and back of the brain and are involved in the processing of sensory information from the body, such as touch and proprioception. The occipital lobes are located at the back of the brain and are primarily involved in the processing of visual information.
Therefore, The correct answer is A i.e Frontal lobes are important for decision-making and judgment.
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This problem involves the same compound that was examined in Problems 6-14 through 6-17, which in the vapor phase was described by the EOS:
with a = 0.3 × 10−16 m3/mol Pa2 K.
This compound has vapor pressures of Psat = 5 50 kPa at T = 323 K and P = 500 kPa at T = 373 K. Saturated liquid at both T = 323 K and T = 373 K has V ≈ 1.25 × 10−4 m3/mol. Estimate each of the following.
A. The fugacity in the vapor phase at T = 323 K and P = 10 kPa.
B. The fugacity in the liquid phase at T =323 K and P = 10 MPa.
C. The fugacity in the vapor phase at T = 373 K and P = 300 kPa.
D. The fugacity in the liquid phase at T = 373 K and P = 10 MPa.
Answers
Answer:
A. The fugacity in the vapor phase at T = 323 K and P = 10 kPa:
Explanation:
QUESTION 5 (Start on a new page.) A block of mass 4 kg starting from rest, at point A, slides down an inclined plane of length 3 m as shown in the diagram below. The plane is inclined by an angle of 30° to the ground. The coefficient of kinetic friction (p) is 0,2 on the inclined plane 5.2 4 kg 5.3 3 m. 30 At the bottom of the inclined plane, at point B, the object slides along a rough horizontal surface experiencing a kinetic frictional force of 19.6 N until it comes to rest at point C 5.1 B State the work-energy theorem in words. Draw a labelled free-body diagram for the block as it slides down the incline. Calculate the: 5.3.1 Kinetic frictional force the block experiences on the incline 5.3.2 Magnitude of the velocity of the block at point B 5.3.3 Distance that the object will slides on the rough horizontal surface until it stops (2) (3) (4) (5) (4) [18]
Answers
1 Therefore, the kinetic frictional force experienced by the block on the incline is 6.784 N.
2 The magnitude of the velocity of the block at point B is approximately 5.11 m/s.
How to calculate the value1. The formula for the kinetic frictional force is given by f = μN, where μ is the coefficient of kinetic friction and N is the normal force. Since the block is on an incline, the normal force can be calculated as N = mg * cos(θ), where θ is the angle of inclination.
N = 4 kg * 9.8 m/s² * cos(30°) = 33.92 N
f = 0.2 * 33.92 N
= 6.784 N
2. Potential energy at point A = mgh, where h is the vertical height of the incline.
Potential energy at point A = 4 kg * 9.8 m/s² * 3 m * sin(30°)
= 58.8 J
The work done by friction is given by W = f * d, where d is the distance traveled along the incline (3 m).
Work done by friction = 6.784 N * 3 m = 20.352 J
Since the work done by friction is negative (opposite to the direction of motion), the total work done on the block is:
Total work = Potential energy at A - Work done by friction
Total work = 58.8 J - 20.352 J = 38.448 J
According to the work-energy theorem, this work done on the block is equal to the change in its kinetic energy. Therefore, we have:
38.448 J = 0.5 * 4 kg * B²
Solving for B, we find:
B = √(38.448 J / (0.5 * 4 kg)) ≈ 5.11 m/s
Therefore, the magnitude of the velocity of the block at point B is approximately 5.11 m/s.
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An 8.0 Kg mass is placed at = 3 where should a 10 Kg mass be placed along the − so that the center of mass will be located ay = 4.5?
Answers
Answer:
Therefore, the 10 kg mass should be placed at x = 5.7 m along the x-axis to achieve a center of mass located at y = 4.5 m.
Explanation:
To find the position along the x-axis where a 10 kg mass should be placed such that the center of mass is located at y = 4.5, we can use the formula for the center of mass:
x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)
Here, m1 and x1 represent the mass and position of the 8 kg mass, respectively. m2 is the mass of the 10 kg mass, and we need to find x2, its position.
Given:
m1 = 8 kg
x1 = 3 m
x_cm = unknown (to be found)
m2 = 10 kg
y_cm = 4.5 m
Since the center of mass is at y = 4.5, we only need to consider the y-coordinate when calculating the center of mass position along the x-axis.
To solve for x2, we can rearrange the formula as follows:
x2 = (x_cm * (m1 + m2) - m1 * x1) / m2
Substituting the given values:
x2 = (x_cm * (8 kg + 10 kg) - 8 kg * 3 m) / 10 kg
Simplifying:
x2 = (x_cm * 18 kg - 24 kg*m) / 10 kg
Now, we can set the y-coordinate of the center of mass equal to 4.5 m and solve for x_cm:
4.5 m = (8 kg * 3 m + 10 kg * x2) / (8 kg + 10 kg)
Simplifying:
4.5 m = (24 kg + 10 kg * x2) / 18 kg
Multiplying both sides by 18 kg:
81 kg*m = 24 kg + 10 kg * x2
Subtracting 24 kg from both sides:
10 kg * x2 = 81 kg*m - 24 kg
Dividing both sides by 10 kg:
x2 = (81 kg*m - 24 kg) / 10 kg
Simplifying:
x2 = 8.1 m - 2.4 m
x2 = 5.7 m
(brainlest?) ples:(
Answer:
the 10 kg mass should be placed at x = -2.4 m to achieve a center of mass at y = 4.5 m.
Explanation:
To find the position along the x-axis where the 10 kg mass should be placed so that the center of mass is located at y = 4.5, we can use the principle of the center of mass.
The center of mass of a system is given by the equation:
x_cm = (m1x1 + m2x2) / (m1 + m2),
where x_cm is the x-coordinate of the center of mass, m1 and m2 are the masses, and x1 and x2 are the positions along the x-axis.
Given:
m1 = 8 kg,
x1 = 3 m,
m2 = 10 kg,
y_cm = 4.5 m.
To solve for x2, we need to find the x-coordinate of the center of mass (x_cm) by using the y-coordinate:
y_cm = (m1y1 + m2y2) / (m1 + m2),
where y1 and y2 are the positions along the y-axis.
Rearranging the equation and substituting the given values:
4.5 = (83 + 10y2) / (8 + 10).
Simplifying the equation:
4.5 = (24 + 10*y2) / 18.
Multiplying both sides by 18:
81 = 24 + 10*y2.
Rearranging the equation:
10*y2 = 81 - 24,
10*y2 = 57.
Dividing both sides by 10:
y2 = 5.7.
Therefore, the y-coordinate of the 10 kg mass should be 5.7 m.
To find the x-coordinate of the 10 kg mass, we can use the equation for the center of mass:
x_cm = (m1x1 + m2x2) / (m1 + m2).
Substituting the given values:
x_cm = (83 + 10x2) / (8 + 10).
Since the center of mass is at x_cm = 0 (the origin), we can solve for x2:
0 = (83 + 10x2) / (8 + 10).
Rearranging the equation:
83 + 10x2 = 0.
24 + 10*x2 = 0.
10*x2 = -24.
Dividing both sides by 10:
x2 = -2.4.
Water flows at a speed of 13 m/s through a pipe that has a diameter of 1.2 m. What is the
diameter of the smaller end of the pipe that the water comes out with a speed of 30 m/s?
Answers
The diameter of the smaller end of the pipe is approximately 0.78 meters.
To determine the diameter of the smaller end of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water should remain constant throughout the pipe.
The mass flow rate is given by the equation:
Mass flow rate = density of water * cross-sectional area * velocity
Since the density of the water remains constant, we can write:
Cross-sectional area1 * velocity1 = Cross-sectional area2 * velocity2
Given that the velocity1 is 13 m/s, the diameter1 is 1.2 m, and the velocity2 is 30 m/s, we can solve for the diameter2 using the equation:
(pi * (diameter1/2)^2) * velocity1 = (pi * (diameter2/2)^2) * velocity2
Simplifying the equation:
(1.2/2)^2 * 13 = (diameter2/2)^2 * 30
Calculating the equation:
(0.6)^2 * 13 = (diameter2/2)^2 * 30
0.36 * 13 = (diameter2/2)^2 * 30
4.68 = (diameter2/2)^2 * 30
Dividing both sides by 30:
0.156 = (diameter2/2)^2
Taking the square root of both sides:
0.39 = diameter2/2
Multiplying both sides by 2:
0.78 = diameter2
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The decibel rules of thumb can be combined. (a) If a sound has intensity xdB, how many dB does a sound 100 times more intense have? (b) If another sound has intensity ydB, how many dB does a sound 4 times less intense have? (c) Combine what you know about (a) and (b): If a sound has intensity zdB, how many dB does a sound 25 times more intense have?
Answers
A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.
Assume: The mass of the atomic nucleus is
about 11.2 the mass of the neutron.
What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?
017 (part 2 of 2) 10.0 points
If the initial kinetic energy of the neutron is
6.58 × 10−13 J, find its final kinetic energy.
Answer in units of J.
Answers
Answer:
The energy of the system and total momentum are both conserved in an elastic head-on collision. The atomic nucleus receives 1.87 percent of the neutron's kinetic energy and final kinetic energy is 12.3×10-13J.
Explanation:
In elastic head-on collision, the energy of the system and total momentum is conserved.
For the neutron,
m - mass, - initial velocity, - final velocity
For the atom,
M- mass , - initial velocity, - final velocity
Conserved momentum on the head-on collision
mv_i=mv_f+MV_f
v_i-v_f=MmV_f
v_i-v_f=14.9V_f.........(I)
The kinetic energy
v_i^2-v_f^2=14.9V_f^2............(II)
From equation (I) and (II)
(v_i-v_f)/(v_i^2-v_f^2 )=(12v_f^2)/(12v_f^2 )
v_i+v_f=V_f.........(III)
Solve equation (I) and (III) for
v_i=7.9v_f
Now, the fraction of the neutron's kinetic energy transferred to the atomic nucleus,
(1/2 MV_f^2)/(1/2 mv_i^2 )=14.9/7.95=1.87
K2=1.87×6.58×10-13J=12.3×10-13J
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lectric Charges
The glass is attracting the pieces of paper. What does
this tell you about the charges on the glass and the
paper?
The glass and the paper have the same charge.
O The glass and the paper have different charges.
ONeither the glass nor the paper has a charge.
Answers
a stone dropps 7,11m how long will it take it to fall
Answers
The time it takes the stone to fall from a height of 7.11 m is 1.2 seconds.
What is time?Time can be defined as an ongoing and continuous sequence of events that occur in succession, from past through the present, and to the future.
To calculate the time it takes the stone to drop from an height of 7.11 m, we use the formula below.
Formula:
H = ut+gt²/2............ Equation 1Where:
H = Heightu = Initial velocityt = Timeg = Acceleration due to gravityFrom the question,
Given:
u = 0 m/sH = 7.11 mg = 9.8 m/s²Substitute these values into equation 1 and solve for t.
7.11 = (0×t)+9.8×t²/27.11 = 4.9t²t² = 7.11/4.9t² = 1.451t = √1.451t = 1.2 secondsHence, the time it takes the stone to fall is 1.2 seconds.
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10. Explain why a power plant doesn’t make electricity.
Answers
Answer: The power station converts stored chemical energy in the fuel to kinetic energy in the raised temperature of the steam that pushed a turbine to rotate the generator that makes electricity.
So chemical to thermal (kinetic) to thermal to kinetic to electrical. Then mostly to heat or movement.
Loads of losses, thermal, mostly.
Explanation: hopes this helps
In all of the power plants, electricity is generated by rotating turbine inside a magnetic field. That is why, power plant transfers mechanical energy into electric energy. That is why, it doesn’t make electricity.
What is power plants?
An industrial facility for the production of electricity is referred to as a power plant. In most cases, power plants are wired into a grid.
A revolving device called a generator, which transforms mechanical power into three-phase electric power, is found in many power plants. An electric current is produced by the movement of a conductor in relation to a magnetic field.
The majority of power plants throughout the world produce electricity by burning fossil fuels including coal, oil, and natural gas. Nuclear energy and the usage of renewable energy sources including solar, wind, geothermal, and hydroelectricity are examples of low-carbon power sources.
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Pls tell me
Question: 1 way to keep people safe from hurricane.
Answers
Answer:
Hurricanes move fast so what you want to do is get into a basement or somthing underground so it will be harder for the hurricane to hit you
Explanation:
thanks for the coins
PLEASE HELP ME! LIKE ASAP! Imagine a population of bugs that has two traits for body color. Some bugs are bright. Some are dark. A new predator can see the bright bugs more easily than the dark bugs. Describe how natural selection could affect this trait in the bug population over time.
Answers
Answer: if the predator sees the light bugs easier than the dark bugs then the bright bug will most likely go extinct
Explanation:
Consider a double-paned window consisting of two panes of glass, each with a thickness of 0.500 cm and an area of 0.760 m2 , separated by a layer of air with a thickness of 1.65 cm . The temperature on one side of the window is 0.00 ∘C; the temperature on the other side is 23.0 ∘C. In addition, note that the thermal conductivity of glass is roughly 36 times greater than that of air. Approximate the heat transfer through this window by ignoring the glass. That is, calculate the heat flow per second through 1.65 cm of air with a temperature difference of 23.0 ∘C . (The exact result for the complete window is 24.4 J/s .)
Answers
The approximate heat transfer through 1.65 cm of air with a temperature difference of 23.0 °C is approximately 24.4 J/s.
To approximate the heat transfer through the air layer in the double-paned window, we can assume that the glass layers have a negligible impact on the heat flow. The heat transfer can be calculated using Fourier's Law of Heat Conduction, which states that the heat flow (Q) is proportional to the temperature difference (ΔT) and inversely proportional to the thickness (L) and thermal conductivity (k) of the material.
First, we need to calculate the effective thermal conductivity of the air layer due to its thickness and the thermal conductivity ratio between air and glass. Let's denote the thermal conductivity of air as k_air and the thermal conductivity of glass as k_glass. Since glass has a thermal conductivity roughly 36 times greater than air, we have k_glass = 36 * k_air.
Next, we calculate the effective thermal conductivity of the air layer as:
k_eff = (k_air * L_air) / (L_air + k_glass)
Substituting the given values, we have:
k_eff = (k_air * 0.0165 m) / (0.0165 m + 0.005 m) = 0.01309 * k_air
Now, we can calculate the heat flow per second through the air layer using the formula:
Q = (k_eff * A * ΔT) / L_air
Substituting the given values, we get:
Q = (0.01309 * k_air * 0.760 m^2 * 23.0 K) / 0.0165 m = 24.4 J/s
Therefore, the approximate heat transfer through 1.65 cm of air with a temperature difference of 23.0 °C is approximately 24.4 J/s.
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