Answers
Answer:
e) 1.04
Explanation:
Related Questions
water flows into the house by means of pipe.inner diameter 2,4cm
absolute pressure 400kpa
flow rate 6
height 4
upper level inner diameter 1,2cm
calculate pressure at the upper level
Answers
The pressure at the upper level of a water flow into the house by means of pipe is 1081 kPa.
How to determine pressure?Calculate the cross-sectional area of the lower pipe:
A₁ = πr₁²
where:
A₁ = cross-sectional area of the lower pipe (m²)
π = mathematical constant (3.14)
r₁ = radius of the lower pipe (m)
A₁ = π(0.12 m)² = 0.0452 m²
Calculate the cross-sectional area of the upper pipe:
A₂ = πr₂²
where:
A₂ = cross-sectional area of the upper pipe (m²)
π = mathematical constant (3.14)
r₂ = radius of the upper pipe (m)
A₂ = π(0.06 m)² = 0.0113 m²
Calculate the flow rate per unit area:
q = Q/A
where:
q = flow rate per unit area (m³/s)
Q = flow rate (m³/s)
A = cross-sectional area (m²)
q = 6 m³/s / 0.0452 m² = 13.28 m²/s
Calculate the velocity of the water in the lower pipe:
v₁ = q/A₁
where:
v₁ = velocity of the water in the lower pipe (m/s)
q = flow rate per unit area (m³/s)
A₁ = cross-sectional area of the lower pipe (m²)
v₁ = 13.28 m²/s / 0.0452 m² = 29.3 m/s
Calculate the velocity of the water in the upper pipe:
v₂ = q/A₂
where:
v₂ = velocity of the water in the upper pipe (m/s)
q = flow rate per unit area (m³/s)
A₂ = cross-sectional area of the upper pipe (m²)
v₂ = 13.28 m²/s / 0.0113 m² = 117.0 m/s
Calculate the head loss:
hL = (v₁² - v2₂²) / 2g
where:
hL = head loss (m)
v₁ = velocity of the water in the lower pipe (m/s)
v₂ = velocity of the water in the upper pipe (m/s)
g = acceleration due to gravity (9.8 m/s²)
hL = (29.3 m/s)² - (117.0 m/s)² / 2(9.8 m/s²) = 23.2 m
Calculate the pressure at the upper level:
p₂ = p₁ + ρghL
where:
p₂ = pressure at the upper level (Pa)
p₁ = pressure at the lower level (Pa)
ρ = density of water (1000 kg/m³)
g = acceleration due to gravity (9.8 m/s²)
hL = head loss (m)
p₂ = 400 kPa + 1000 kg/m³(9.8 m/s²)(23.2 m) = 1081 kPa
Therefore, the pressure at the upper level is 1081 kPa.
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On March 27, 2004, the United States successfully tested the hypersonic X-43A scramjet, which flew at Mach 7.0 (seven times the speed of sound) for 11 seconds. (A scramjet gets its oxygen directly from the air, rather than from fuel.) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Swim competition. Part A At this rate, how many minutes would it take such a scramjet to carry passengers the approximately 5000 kmkm from San Francisco to New York? (Use 331 m/sm/s for the speed of soun
Answers
Answer:
Explanation:
Speed of sound = 331 m /s
speed of jet = 7 .00 Mach = 7 times speed of sound
= 7 x 331 = 2317 m /s
distance to be covered = 5000 x 1000 = 5 x 10⁶ m
Time taken = distance / speed of jet
= 5 x 10⁶ / 2317
= 2.158 x 10³ s
= 35.96 minutes .
If Earth were a perfect sphere, would you weigh more or less at the equator than at the poles? Explain
Answers
Answer:
You would weigh the same.
Explanation:
At the moment, since Earth is not a perfect sphere, the Earth "bulges out" at the equator, so you're further from the centre of the Earth. Since gravity acts through a body's center of mass, the further you are from the centre the weaker the gravitational acceleration you will feel, because gravity weakens over distance.
So, you're actually lighter at the equator than you'd be at the poles.
However, if the Earth was a perfect sphere, this "bulge" at the equator would not happen, and so you would weigh the same at the poles and at the equator.
Hope this makes sense.
Problem Try to answer the following questions:(a) What is the maximum height above ground reached by the ball?(b) What are the magnitude and the direction of the velocity of the ball just before it hits the ground? Show Your Problem Solving Steps: Show these below:1) Draw a Sketch2) Choose origin, coordinate direction3) Inventory List – What is known?4) Write the kinematics equation(s) and solution of Part (a):5) Write the kinematics equation(s) and solution of Part (b):Problem 3 A small ball is launched at an angle of 30.0 degrees above the horizontal. It reaches a maximum height of 2.5 m with respect to the launch position. Find (a) the initial velocity of the ball when it’s launched and (b) its range, defined as the horizontal distance traveled until it returns to his original height. As always you can ignore air resistance.(a) Initial velocity [Hints: How is v0 related to vx0 and vy0. How can you use the information given to calculate either or both of the components of the initial velocity?](b) Range [Hints: This problem is very similar to today’s Lab Challenge except that for the challenge the ball will land at a different height.]
Answers
3.
\(\begin{gathered} \theta=30^{\circ} \\ y_{\max }=2.5m \end{gathered}\)a)
\(\begin{gathered} y_{\max }=\frac{v^2\sin ^2(\theta)}{g} \\ \end{gathered}\)Solve for v:
\(\begin{gathered} v=\sqrt[]{\frac{y_{\max }\cdot g}{\sin ^2(\theta)}} \\ v=98\cdot\frac{m}{s} \end{gathered}\)b)
\(\begin{gathered} r=\frac{v^2}{9}\sin (2\theta) \\ r=\frac{98^2}{9.8}\cdot\sin (2\cdot30) \\ r=\frac{98^2}{9.8}\sin (60) \\ r=848.7m \end{gathered}\)I need the fine the magnitude of the vector sum. ( please see photo )
Answers
In order to find the magnitude of the vector sum, proceed as follow:
First, take into account that the vertical and horizontal components of A are (by considering each grid square is 9.00N):
Ax = 3*9.00N = 27.00N
Ay = 4*9.00N = 36.00N
Now, consider that B vector only has a vertical component:
By = -4*9.00N = -36.00N
and the C vector only has a horizontal component:
Cx = -2*9.00N = -18.00N
By and Cx are negative because these component are in the negative part of the y and x axis respectively.
Now, simplify all vertical and horizontal components. It determines the x and y components of the sum vector S:
Sx = Ax+Cx = 27.00N - 18.00N = 9.00N
Sy = Ay+By = 36.00N - 36.00N = 0.00N
Finally, the magnitude of the sum vector is:
\(S=\sqrt[]{(S_x)^2+(S_y)^2}=\sqrt[]{(9.00N)^2+(0.00N)^2}=9.00N\)The magnitude is 9.00N
Question
In order for work to be done, what three things are necessary
Answers
A 24.4kg dog is running northward at 2.14m/s, while a 5.53kg cat is running eastward at 3.56m/s. Their 78.5kg owner has the same momentum as the two pets taken together. Find the direction of the owner's velocity. Find the magnitude of the owner's velocity.
Answers
The owner's velocity is in the opposite direction of the combined velocity of the dog and the cat, and its magnitude is approximately 0.916 m/s.
To solve the given problem, we can use the principle of conservation of momentum to find the direction and magnitude of the owner's velocity.
Let's denote the velocity of the dog as v1 (northward), the velocity of the cat as v2 (eastward), and the velocity of the owner as v (unknown).
According to the conservation of momentum, the total momentum before the interaction is equal to the total momentum after the interaction.
The total momentum before the interaction is given by:
Total momentum before = (mass of the dog * velocity of the dog) + (mass of the cat * velocity of the cat) + (mass of the owner * velocity of the owner)
Mass of the dog (m1) = 24.4 kg
Velocity of the dog (v1) = 2.14 m/s
Mass of the cat (m2) = 5.53 kg
Velocity of the cat (v2) = 3.56 m/s
Mass of the owner (m3) = 78.5 kg
Velocity of the owner (v) = unknown
Total momentum before = (24.4 kg * 2.14 m/s) + (5.53 kg * 3.56 m/s) + (78.5 kg * v)
The total momentum after the interaction is zero since the owner has the same momentum as the pets taken together.
Total momentum after = 0
Equating the two expressions:
(24.4 kg * 2.14 m/s) + (5.53 kg * 3.56 m/s) + (78.5 kg * v) = 0
Simplifying the equation:
(52.216 kg·m/s) + (19.6488 kg·m/s) + (78.5 kg * v) = 0
71.8648 kg·m/s + (78.5 kg * v) = 0
Solving for v:
78.5 kg * v = -71.8648 kg·m/s
v = -71.8648 kg·m/s / 78.5 kg
v ≈ -0.916 m/s
Therefore, the direction of the owner's velocity is opposite to the combined velocity of the dog and the cat, and the magnitude of the owner's velocity is approximately 0.916 m/s.
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Two forces F1 = -7.50i + 4.90j and F2 = 8.20i + 3.50j are acting on an object with a mass of m = 2.70 kg. The forces are measured in newtons, i and j are the unit vectors. What is the magnitude of the object's acceleration?
Answers
The magnitude of the object's acceleration is determined as 3.12 m/s².
What is the resultant force on the object?The magnitude of resultant force acting on the object is calculated as follows;
F = √[(x₁i + x₂i)² + (y₁j + y₂j)²]
where;
x₁i is the initial force in x-directionx₂i is the final force in x-directiony₁j is the initial force in y-directiony₂j is the final force in y-directionF = √[(-7.5 + 8.2)² + (4.9 + 3.5)²]
F = 8.43 N
The magnitude of the object's acceleration is calculated by applying Newton's second law of motion as follows;
F = ma
a = F/m
where;
F is the applied force on the objectm is the mass of the objecta is the acceleration of the objecta = (8.43) / (2.7)
a = 3.12 m/s²
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D. O3.015 m/sE. O 9.381 m/s7. An object of mass 18 kg moving with a speedof 23 m/s to the right collides with an objectof mass 17 kg moving with a speed of 16 m/sin the same direction. After collision, the 18 kg object movesto the right with a speed of 3 m/s to the right. Calculatethe velocity of the 17 kg object after collision. (1 point)A.40.894 m/sB. O 44.612 m/sC. O 26.024 m/sD.37.176 m/sAE. O 22.306 m/s
Answers
We are given the following information
Before collision:
Mass of the 1st object = 18 kg
Initial speed of the 1st object = 23 m/s
Mass of the 2nd object = 17 kg
Initial speed of the 2nd object = 16 m/s
After collision:
Final speed of the 1st object = 3 m/s
Final speed of the 2nd object = ?
Recall from the law of conservation of momentum, the total momentum before the collision and after the collision must be equal.
\(\begin{gathered} momentum\;before=momentum\;after \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \end{gathered}\)Let us substitute the given values and solve for the final speed of the 2nd object (v2).
\(\begin{gathered} 18\cdot23+17\cdot16=18\cdot3+17\cdot v_2 \\ 686=54+17\cdot v_2 \\ 17\cdot v_2=686-54 \\ v_2=\frac{632}{17} \\ v_2=37.176\;\;\frac{m}{s} \end{gathered}\)Therefore, the velocity of the 17 kg object after the collision is 37.176 m/s
Option D is the correct answer.
how much power an appliance uses
Answers
Answer:
it depends on the appliance
Explanation:
bigger appliances will use more power, smaller appliances will use a lesser amount
Three cars (car F, car G, and car H) are moving with the same velocity when the driver suddenly slams on the brakes, locking the wheels. The most massive car is car F, the least massive is car H, and all three cars have identical tires. (a) Which car travels the longest distance to skid to a stop? A) Car F B) Car G C) Car H D) They all travel the same distance in stopping. (b) For which car does friction do the largest amount of work in stopping the car? A) Car F B) Car G C) Car H D) The amount of work done by friction is the same for all cars.
Answers
Answer:
a) Option D
b) Option A
Explanation:
a) Option D
Because a massive car will have more inertia which will make the car move faster but a massive car simultaneously will have more friction thereby restricting its movement in the forward direction. Hence, all the three cars will move equal distance.
b) Option A, Car F
Being most massive car, the frictional force required to stop the car will be highest.
A box-shaped metal can has dimensions 5 in. by 19 in. by 4 in. high. All of the air inside the can is removed with a vacuum pump. Assuming normal atmospheric pressure outside the can, find the total force (in lb) on one of the 5-by-19 in. sides. (Enter the magnitude.)
Answers
Answer:
The force is \(F = 1397 lb\)
Explanation:
From the question we are told that
The length of the box is \(l = 19 \ in\)
The width of the box is \(w = 5 \ in\)
The height is \(h = 4\ in\)
The pressure experience on one of the sides is mathematically represented as
\(p = \frac{F}{A}\)
Where A is the area of the box which is mathematically evaluated as
\(A = l * w\)
substituting values
\(A = 5 *19\)
\(A = 95 \ in^2\)
This pressure is equivalent to the atmospheric pressure which has a constant value of \(p = 14.7 pi\)
This implies that
\(14.7 = \frac{F}{95}\)
=> \(F = 14.7 *95\)
=> \(F = 1397 lb\)
A man walks 8 m east in 12 seconds . What is the man's velocity ?
Answers
Answer:
0.67m/s due east
Explanation:
Given parameters:
Displacement of the man = 8m east
Time taken = 12s
Unknown:
Velocity of the man = ?
Solution:
The velocity of a body is the rate of displacement per time;
Velocity = \(\frac{Displacement}{time}\)
Velocity = \(\frac{8}{12}\) = 0.67m/s due east
Select the correct answer.
A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward point A. What are the total
distance and total displacement of the boat?
O A.
OB.
O C.
O D.
The total distance is 105 kilometers and the total displacement is 45 kilometers east.
The total distance is 60 kilometers and the total displacement is 60 kilometers east.
The total distance is 105 kilometers and the total displacement is 15 kilometers east.
The total distance is 60 kilometers and the total displacement is 45 kilometers east.
Answers
The total distance is 105 kilometers and the total displacement is 15 kilometers east. Option C
How to solve for the total distanceTo calculate the total distance, we add the distances traveled in each leg of the journey: 60 kilometers (from A to B) + 45 kilometers (from B back to A) = 105 kilometers.
However, displacement refers to the change in position of an object in a straight line from its starting point to its ending point. In this case, since the boat starts and ends at the same point (A), the total displacement is zero.
Hence The total distance is 105 kilometers and the total displacement is 15 kilometers east.
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A wall clock has a second hand 25.0 cm long.
For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Fast car, flat curve.
Part A
What is the radial acceleration of the tip of this hand?
Express your answer in centimeters per second squared.
Answers
The radial acceleration of the tip of this hand is 0.274 cm/s². The answer is expressed in centimeters per second squared.
What is acceleration?The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The radial acceleration of the tip of this hand is found as;
\(\rm \omega = \frac{2 \pi}{60} \\\\ a_r= \omega^2r\\\\ a= (\frac{2\pi}{60} )^2\times 25\\\\a=0.274 \ cm/s^2\)
Hence,the radial acceleration of the tip of this hand is 0.274 cm/s²
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What is the average acceleration of a me
that reaches 24 m/s in 5.5 seconds?
F 0.20 m/s2
G 2.5 m/s
H 10 m/s2
J 208 m/s
Answers
Answer:
See below
Explanation:
Assuming starting from zero velocity
vfinal = at
24 = a (5.5)
a = 4.36 m/s^2 I think part of your question is missing....what was the initial velocity?
what are some possible results of constructive forces on earth's surface
Answers
Answer:
what are the choices
Explanation:
Answer:
Landforms are a result of a combination of constructive and destructive forces. Collection and analysis of data indicates that constructive forces include crustal deformation, faulting, volcanic eruption and deposition of sediment, while destructive forces include weathering and erosion.
Explanation:
A net force F acts on a mass m and produces an acceleration a. What acceleration results if a net force 4F acts on a mass 6m?
Answers
Answer: The acceleration results if a net force of 4F acts on a mass of 6m is 2/3a.
Explanation:
Force exerted on an object can be defined as a pull or a push on an object which leads to it's displacement. Force is taken to be a vector quantity because it has both magnitude and direction. The formula which can be used to determine force exerted on an object in physics is:
F= mass( kg) × acceleration( m/ s²)
Acceleration is defined as the rate at which the velocity of an object changes. From the formula of force given above it can be determined by making it the subject of formula. Therefore acceleration= Force/ mass.
From the question,
Force= 4F
Mass= 6m
Therefore acceleration= F/m
= 4/6
Acceleration= 2/3a
The required magnitude of acceleration when force is 4F and mass is 6m is 2/3a.
Given data:
The magnitude of net force is, 4F.
The value of mass is, 6m.
Apply the Newton's second law which says that force exerted on an object can be defined as a pull or a push on an object which leads to it's displacement. Force is taken to be a vector quantity because it has both magnitude and direction. The formula which can be used to determine force exerted on an object in physics is
F = ma
a = F/m ..........................................(1)
here, a is the acceleration.
Solving as when the force becomes 4F and mass becomes 6m.
\((4F) = (6m) \times a'\\\\a '= \dfrac{4F}{6m}\\\\a '= \dfrac{2}{3}a\)
Thus, the required magnitude of acceleration when force is 4F and mass is 6m is 2/3a.
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3/4+ (10 * 7/3- 4) * (-11) -2
Answers
Answer:
-210.25
Explanation:
you should start at the braces, and start with multipe the numbers and then have the answer
Sequence of operations:
1.Brackets
2.braces
3.multiply
4. plus
5.negetive
Collision practice
Running at 2.0 m/s, Bruce, the 45.0-kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the other direction. Upon collision,
Biff wraps up Biff for the tackle. What is the resultant velocity of the players.
V=————— m/s
Answers
Answer:
\(v_f=4\:\mathrm{m/s\: in\:Biff's\:direction}\)
Explanation:
Using the Law of Conservation of Momentum, we can set up the following equation:
\(m_2v_2-m_1v_1=m_fv_f\), where \(m_1v_1\) is Bruce's momentum and \(m_2v_2\).
Plugging in given values, we get:
\(90\cdot7-45\cdot2=(90+45)v_f, \\v_f=\frac{540}{135}=\fbox{$4\:\mathrm{m/s}$}\).
When will sky diver reach terminal velocity after he has opened his parachute?
Answers
Answer:
As the skydiver gains speed, their weight stays the same but the air resistance increases. There is still a resultant force acting downwards, but this gradually decreases. Eventually, the skydiver's weight is balanced by the air resistance. There is no resultant force and the skydiver reaches terminal velocity.
the distance that a body in free fall falls each second is
a. is about 9.8m
b. is about 19.6m
c. increases as time passes
d. none of the above are correct
Answers
The distance that a body in free fall falls each second is (A). is about 9.8m is correct option.
The distance that a body in free fall falls each second is about 9.8 meters, which is equivalent to the acceleration due to gravity on the Earth's surface.
This means that in the absence of air resistance, an object dropped from a certain height will fall 9.8 meters during the first second, 19.6 meters during the second second, 29.4 meters during the third second, and so on. The distance increases as time passes, but not at a constant rate, as it is being affected by the acceleration due to gravity.
Therefore, the correct option is (a) .
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A man walks 30 m to the west, then 5 m to the east in 45 seconds.
What is his average speed?
Answers
A 10.0-kg microwave oven is pushed 8.00 m up the sloping surface of a loading ramp inclined at an angle of above the horizontal, by a constant force \vec{F} F with a magnitude 110 N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250. (a) What is the work done on the oven by the force ? (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven’s kinetic energy. (e) Use \Sigma \vec{\boldsymbol{F}}=m \vec{\boldsymbol{a}}Σ F =m a to calculate the acceleration of the oven. Assuming that the oven is initially at rest, use the acceleration to calculate the oven’s speed after traveling 8.00 m. From this, compute the increase in the oven’s kinetic energy, and compare it to the answer you got in part (d)
Answers
(a) Work done on the oven = 110J
(b) Work done by the friction force =250J
(c) Increase in PE =1000J
(d) Increase in KE = 640J
(e) Acceleration of the oven =14.9 m/s² And KE =74.5J
Given
g = 10
mass = 10kg
distance = 8 m
Magnitude of force = 110 N
Kinetic friction = 0.250 N
A
The work done on the oven by the force is given by
W = F.d
W = 110*14cos(0)
It is a Horizontal component
W = 110 J
B
Now to calculate the work done by the friction force
Frictional force = μ * N
Frictional force = 0.25*10*10
Frictional force = 25 N
Frictional work = frictional force . d
work(f) = 25 * 10cos(0)
Work(f) = 250 J
C
Increase in the Potential energy
PE = mgh
PE = 10*10*10cos(0)
PE = 1000 J
D
The increase in the oven's kinetic energy
PE₁ = KE₁ + W = PE₂ + KE₂
110 - 260 - 1000 = KE₂
KE₂ = 640J
E
The acceleration of the oven
Force - frictional force = ma
a = 110 -250/10 = 14 m/s²
Now to calculate KE we should know the velocity
V(f)² = v(i)² + 2ad
v(f)² = 2(14)(8) = 224
v(f) = 14.9 m/s
Now KE will be given as
(1/2)mv²
KE = (1/2)(10)(14.9)
KE = 74.5 J
Therefore, we got to know the work done, PE and KE just by the force applied.
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How did the approximate speed
of the marbles belore each collision compare
to after each collision?
Answers
Answer:
i think this may help you
Explanation:
In an inelastic collision the total kinetic energy after the collision is not equal to the total ... If there are no net forces at work (collision takes place on a frictionless surface ... Momentum is equal to the product of mass and velocity. ... bodies before the collision is equal to the total kinetic energy of the bodies after the collision.
Which statements accurately describe sound waves? Check all that apply.
Sound waves are transverse waves.
Sound waves require a medium to transfer energy.
Sound is heard when a vibration strikes the ear.
Sound waves can only travel through liquids and gases.
When particles of a medium interact, part of the wave’s energy is lost.
A wave’s energy can be distinguished from other movements of the medium.
Answers
The statements that accurately describe sound waves are:
2,3,4,6
1. Sound waves require a medium to transfer energy. Unlike electromagnetic waves, such as light, sound waves cannot propagate through a vacuum. They need a material medium, such as air, water, or solids, to transfer their energy.
2. Sound is heard when a vibration strikes the ear. Sound is a mechanical wave that is produced by vibrations or oscillations of objects. When these vibrations reach our ears, they are detected by the auditory system, which allows us to perceive sound.
3. When particles of a medium interact, part of the wave's energy is lost. Sound waves experience energy losses due to factors like friction, absorption, and scattering. As the wave propagates through a medium, some of its energy is converted into other forms, such as heat, resulting in a decrease in the wave's intensity.
4. A wave's energy can be distinguished from other movements of the medium. Sound waves carry energy in the form of vibrations or oscillations of particles within a medium. These movements are distinct from other random or uncorrelated motions of the medium's particles that do not contribute to the propagation of the sound wave.
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A⃗ = 22x^ + (-14)y^, and B⃗ = 2.5x^ + 13y^. Express A⃗ −B⃗ in unit vector notation.
Answers
The unit vector notation in the given direction is \(0.58 \hat x + 0.81 \hat y\).
What is vector quantity?A physical quantity that has both directions and magnitude is referred to as a vector quantity.
A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.
In terms of subtracting a vector, it works the same as adding the vector's opposite.
Given that:
\(\vec A = 22 \hat x + (-14)\hat y,\\\vec B = 2.5\hat x + 13\hat y.\)
( two given vectors are written here.)
Therefore: the subtraction of these two vectors will be:
\(\vec A - \vec B = (= 22 \hat x + (-14)\hat y) - (2.5\hat x + 13\hat y)\\= 19.5\hat x - 27 \hat y\)
Unit vector in that direction of vectors is =
\(= \frac{19.5\hat x - 27 \hat y}{\sqrt{19.5^2 + (-27)^2} } \\= 0.58 \hat x + 0.81 \hat y\)
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A body dropped over a fixed rough inclined plane of inclination 45 from height h. If after collision velocity of body becomes horizontal
then co-efficient of restitution if co-efficient of friction is 1/2
Answers
As per the given scenario, in this case, the coefficient of friction () is half and the coefficient of restitution (e) is zero.
Identify the body's starting velocity:
We may use the equation of motion to get the body's initial velocity (u)
\(v^2 = u^2 + 2as\)
\(0 = u^2 + 2(-9.8)m/s^2 * h\)
\(u^2 = 19.6h\)
u = √(19.6h)
Determine the coefficient of restitution (e):The body's initial velocity (u) and initial relative velocity (u_rel) are the same.
The body's horizontal velocity immediately following the collision, which is zero, is the final relative velocity (v_rel).
\(e = v_{rel }/ u_{rel}\)
e = 0 / u_rel = 0 / u
Now, one can investigate the forces affecting the body: When a body is on an inclined plane.
There are two main forces at work on it: the frictional force that prevents the body from moving and the gravitational force that pulls it downward (mg).
The gravitational force has two components that act perpendicular to and parallel to the inclined plane, respectively: m*g*cos(45°) and m*g*sin(45°).
Determine the conditions for the body to stop:
μ * N = m * g * sin(45°)
μ * (m * g * cos(45°)) = m * g * sin(45°)
μ * cos(45°) = sin(45°)
(1/2) * cos(45°) = sin(45°)
Simplifying further, we have:
√2 / 4 = √2 / 2
Thus, the body will come to rest following the collision if the equation is valid.
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A plate of iron at 20 °C has shown in the figure below. If the temperature
raised to 100 °C and the coefficient of linear expansion of iron is 1.1 x 10-7 o
1, then what is the final area of the plate?
(5
2 m
2 m
Answers
The final area of the plate is 4.0000352 \(m^2\) if the temperature raised to 100 °C and the coefficient of linear expansion of iron is 1.1 x 10-7.
Expecting that the plate of iron is rectangular, we can involve the recipe for warm extension of solids to compute the last region of the plate. The equation for direct warm development is given by ΔL = αLΔT, where ΔL is the adjustment of length, α is the coefficient of straight extension, L is the first length, and ΔT is the adjustment of temperature.
Since the region of the plate is given by A = L*W, where L is the length and W is the width, we can involve the equation for straight warm extension to compute the adjustment of length of the plate and afterward use it to compute the last region.
ΔL = αLΔT = \((1.1 x 10^-7 m/oC)(2 m)(80 oC) = 1.76 x 10^-5 m\)
The last length of the plate is L + ΔL = 2 m + 1.76 x \(10^-5\) m = 2.0000176 m (approx.)
The last width of the plate is thought to be unaltered as it isn't impacted by the adjustment of temperature.
Thusly, the last region of the plate is A = L*W = (2.0000176 m)(2 m) = 4.0000352 \(m^2\) (approx.)
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Refer to the picture!
Answers
(a) The work done by the donkey on the cart is 59,721.9 J.
(b) The work done by the force of gravity on the cart is -48,434.87 J.
(c) The work done on the cart by friction during this time is 11,315.12 J.
What is the work done by the donkey on the cart?(a) The work done by the donkey on the cart is calculated as follows;
Wd = Fd cosθ
where;
F is the applied force by the donkeyd is the displacementθ is the angle of inclinationWd = 375 N x 163 m x cos(12.3)
Wd = 59,721.9 J
(b) The work done by the force of gravity on the cart is calculated as;
Wg = Fg x d x cosθ
Where;
Fg is the force of gravityd is the displacementθ is the angle between the force of gravity and displacementθ = 90⁰ + 4.03⁰ = 94.03⁰
Wg = (431 kg x 9.81 m/s²) x 163 m x cos (94.03)
Wg = -48,434.87 J
(c) The work done on the cart by friction during this time is calculated as;
Wf = Ff x d x cosθ
where;
Ff is the force of friction;Ff = μmg cosθ
Ff = 0.0165 x 431 kg x 9.81 x cos (4.03)
Ff = 69.59 N
Wf = 69.59 x 163 x cos (4.03)
Wf = 11,315.12 J
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