A wood block is placed on a rough surface. The surface starts horizontal, and one end is then raised so that the angle the surface makes with the horizontal gradually increases until, at an angle θ , the block begins to move down the surface, as shown in the figure. From this observation, which of the following can be concluded for angles greater than θ ?
A) there is no frictional force on the block.
B) The gravitational force is equal to the frictional force.
C)The component of the gravitational force along the surface is greater than the frictional force.
D)The normal force is equal to the gravitational force.
E )The component of the normal force along the surface is greater than the gravitational force.
Answers
Answer:
The correct option is;
C) The component of the gravitational force along the surface is greater than the friction force.
Explanation:
The force of static friction acts along the plane in opposite direction to the component of the force of the weight acting down (along) the plane.
The magnitude of the frictional force, \(F_{fr}\) is given by the following formula;
\(F_{fr}\) = μ·N = (m·g·cos(θ))·μ
The magnitude of the weight acting along the incline plane = m·g·sin(θ)
As θ increases, sin(θ) increases, and cos(θ) decreases, such that magnitude of the weight acting downwards, along the inclined plane becomes larger than the frictional force acting upwards along the incline plane.
Related Questions
Two litres of water, initially at 20 °C, is heated to 40 °C. Determine thevolume of water at 40 °C if the coefficient of volumetric expansion ofwater within this range is 30 x 10-9 °C.
Answers
Given data
*The given initial volume of water is V_1 = 2 L
*The given initial temperature of the water is T_i = 20 °C
*The given final temperature of the water is T_f = 40 °C
*The coefficient of volumetric expansion of water is
\(\alpha=30\times10^{-9}^{}\text{C}\)The formula for the volumetric expansion is given as
\(\frac{\Delta V}{V}=\alpha(T_f-T_i)\)Substitute the known values in the above expression as
\(\begin{gathered} \frac{\Delta V}{V_1}=30\times10^{-9}\times(40^0-20^0) \\ V_2-V_1=6\times10^{-7}V_1 \\ V_2=6\times10^{-7}V_1+V_1^{} \\ =V_1(6\times10^{-7}+1) \\ =2.00\text{ L} \end{gathered}\)A Carnot refrigeration cycle is used to maintain a room at
23 °C by removing heat from groundwater at 15 °C.
Refrigerant R-134a enters the condenser as saturated
vapor at 40 °C and leaves as saturated liquid at the
same temperature. The evaporator pressure is 351 kPa.
a) If the room is to receive 2kW, what is power input to
the compressor?
b) Net power input to cycle?
Answers
a) The power input to the compressor in the Carnot refrigeration cycle, in order to supply 2 kW of cooling to the room, will depend on the efficiency of the cycle and the heat transfer involved.
b) The net power input to the cycle can be determined by considering the work done by the compressor and the work done on the system.
a) To calculate the power input to the compressor, we need to determine the heat transfer from the groundwater to the room. The Carnot refrigeration cycle is an idealized cycle, and its efficiency is given by the equation: Efficiency = 1 - (T_evaporator / T_condenser), where T_evaporator and T_condenser are the temperatures at the evaporator and condenser, respectively. Using this efficiency, we can calculate the heat transfer from the groundwater and convert it to power input.
b) The net power input to the cycle takes into account the work done by the compressor and the work done on the system. It can be calculated by subtracting the work done by the compressor from the heat transfer from the groundwater. The work done by the compressor can be determined using the power input calculated in part a), and the heat transfer from the groundwater can be obtained using the given temperatures and the specific heat properties of the refrigerant.
Overall, the Carnot refrigeration cycle involves several calculations to determine the power input to the compressor and the net power input to the cycle, considering the heat transfer and work done in the system.
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Two light waves of the same frequency start out in phase (with amplitudes going up at the same moment), and they interfere having traveled different distances. What happens if the path difference in the two waves is 600 nm and the wavelength of the light is 400 nm (blue light)
Answers
The path difference is 600 nm and the wavelength of the light is 400 nm (blue light). Since the path difference (600 nm) is not a multiple of the wavelength (400 nm), the two waves will interfere destructively,
When two waves of the same frequency and amplitude interfere, the resulting wave is determined by the phase difference between them. If the two waves are in phase when they start, then they will continue to be in phase until they encounter a path difference. In this case, the path difference is 600 nm, which is 1.5 times the wavelength of the blue light (400 nm).
When the two waves interfere, they will produce a pattern of interference known as a diffraction pattern. In this case, the path difference is large enough that the two waves will interfere destructively, meaning that the amplitudes of the waves will cancel each other out at certain points along the diffraction pattern. The exact locations of these points depend on the angle of incidence, but in general, they will be spaced at regular intervals corresponding to the wavelength of the light.
Therefore, when two light waves of the same frequency start out in phase and interfere having traveled different distances, if the path difference in the two waves is 600 nm and the wavelength of the light is 400 nm (blue light), the interference will be destructive and result in a diffraction pattern with points of cancellation spaced at regular intervals corresponding to the wavelength of the light.
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como te llamas?
(nom
Answers
Answer: como mi llamo yo?
Explanation: in rlf Maria
Answer:nai
Explanation:
Use the table of electric force between objects in two different interactions to answer the question. Interaction Charge on Object 1 (C) Charge on Object 2 (C) Amount of Force A +1 -2 weak B +1 +2 strong Which conclusion about the interactions is correct? (1 point) O The objects in interaction A are farther apart than in interaction B. O The objects in interaction A are smaller than in interaction B O Positive charges are stronger than negative charges. O Repulsive forces are stronger than attractive forces. Type here to search
Answers
Answer:
The objects in interaction A are farther apart than in interaction B.
The objects in interaction A are farther apart than in interaction B.
What is coulomb's law?According to Coulomb's law: The magnitude of each of the electric forces with which two point-at-rest charges interact is directly proportional to the product of the magnitude of both charges.
And inversely proportional to the square of the distance that separates them and has the direction of the line that joins them.
The formula will be given as:
] \(F=\dfrac{kq_1q_2}{r^2}\)
Intersaction A is having weak force since the magnitude of the force is same in both interaction A and B from the formula of force the force is also dependent on the distance between the charge.
Thus the weak force in the interaction A is due to the higher distance between the two charges.
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What velocity must a 1240 kg car have in order to have the same
momentum as a 3660 kg truck traveling at a velocity of 12 m/s to the
west?
Answers
Answer:
90
Explanation:
cause
When three vectors are added graphically and form a closed triangle, the largest enclosed angle between any two of the vectors cannot be greater than:
a. 600.
b. 900.
c. 1800.
d. No maximum exists.
Answers
The largest enclosed angle between any two vectors in a closed triangle cannot be greater than 180 degrees.
Closed triangle: When three vectors are added graphically and form a closed triangle, it means that the starting point and the ending point of the vector addition form a triangle.
Triangle angles: In a triangle, the sum of the three angles is always equal to 180 degrees. This is a fundamental property of triangles.
Vector addition: When three vectors are added graphically to form a closed triangle, the starting point of the first vector is connected to the ending point of the second vector, and the starting point of the second vector is connected to the ending point of the third vector. This results in a closed triangle.
Enclosed angles: The enclosed angles between the vectors in the closed triangle are the angles between the connected ends of the vectors.
Largest enclosed angle: Since the sum of the angles in a triangle is 180 degrees, the largest enclosed angle between any two vectors in the closed triangle cannot be greater than 180 degrees. This is because if one angle were larger than 180 degrees, the sum of the angles in the triangle would exceed 180 degrees, which is not possible.
Therefore, the largest enclosed angle between any two vectors in a closed triangle cannot be greater than 180 degrees.
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A sportscar has a mass of 1500 kg and accelerates at 5 meters per second squared. What is the magnitude of the force acting on the sportscar?
Answers
Answer:
7500 Newtons
Explanation:
Mass of the sportscar= 1500 kg
Acceleration of the sportscar= 5m/s^2
Hence, let the Force acting on it be F
\(We\ know\ that,\\Force=Mass*Acceleration\\F=ma\\\\Here,\\F=1500*5\\=7500 kg m/s^2\ or\ 7500\ Newtons\)
HELP!!!!!!! if you were to ride a bicycle on a rocky pathway, how much friction would there be and what would be the thermal and friction energy level?
Answers
Answer: The equation for calculating friction is fr=FR/N
Explanation:
Fr (retentive force : friction) divided by N (normal/perpendicular force) equals fr (friction)
Two point charges are located at the following locations:
q1= 2.5 × 10^−5 C located at ~r1= <−4,3,0> m
q2= −5×10^−5C located at ~r2= < 4,−3,0> m.
a) Calculate the net electric force on an electron located at the origin. Answer must be a vector.
b) Determine where to place a positive charge q3= 1.2×10^−5C so that the net force on the electron located at the origin is zero.
Answers
a) The net electric force on an electron located at the origin is Fₑ = <0, 0, 5.4 × 10⁻³> N.
(b) the size of the system is not mentioned, so it is assumed to be small enough that the charges can be treated as point charges.
Determine the net electric force?To calculate the net electric force on the electron, we need to consider the electric forces exerted by each of the point charges. The electric force between two charges is given by Coulomb's law:
F = (k * |q1 * q2|) / r²
where k is the electrostatic constant (k ≈ 8.99 × 10⁹ N m²/C²), q1 and q2 are the charges, and r is the distance between them.
For the first charge (q1), located at position ~r1 = <-4, 3, 0> m, the distance vector between the origin and q1 is r1 = <-4, 3, 0> m.
For the second charge (q2), located at position ~r2 = <4, -3, 0> m, the distance vector between the origin and q2 is r2 = <4, -3, 0> m.
To calculate the net electric force, we sum the individual forces vectorially.
The force exerted by q1 on the electron is directed towards q1, while the force exerted by q2 is directed away from q2. The x and y components of the forces cancel out, while the z component adds up, resulting in a net force of Fₑ = <0, 0, 5.4 × 10⁻³> N.
b) To find the position where a positive charge q₃ = 1.2 × 10⁻⁵ C should be placed so that the net force on the electron at the origin is zero, we need to consider the principle of superposition.
Determine the net force on the electron?The net force on the electron is the vector sum of the forces exerted by q₁, q₂, and q₃.
Since the net force on the electron is zero, the vector sum of the forces must be equal to the negative of the force exerted by q₁ and q₂. Mathematically, this can be represented as:
F₁ + F₂ + F₃ = -Fₑ
where F₁, F₂, and F₃ are the forces exerted by q₁, q₂, and q₃, respectively, and Fₑ is the net electric force calculated in part (a).
To find the position where q₃ should be placed, we need to solve this equation by setting up a system of equations. The coordinates of q₃ can be represented as ~r₃ = <x, y, z> m. By substituting the known values for F₁, F₂, F₃, and Fₑ, we can solve for x, y, and z.
However, please note that the problem does not provide the mass or charge of the electron, which could affect the net force calculation.
Additionally, the size of the system is not mentioned, so it is assumed to be small enough that the charges can be treated as point charges.
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A box is pushed horizontally with constant speed across a rough horizontal surface.
Which of the following must be true?
Answers
Answer:
what is the answers? so i can't answer.
Explanation:
The statement that is true as regards movement of the box when pushed horizontally with constant velocity is given by;
Option D; The friction is equal to the applied force
The missing options are;
a) The friction is less than the applied force
b) There is no friction acting on the box
c) The friction is greater than the applied force
d) The friction is equal to the applied force
We are told that the box is pushed with constant speed. We know that at constant speed, acceleration is equal to zero. If acceleration is zero, the 2 forces acting on the box will be the applied force and the frictional force from the rough horizontal surface.
Now, if the frictional force was less than the applied force, it means the velocity of the box will gradually be increasing but we are told the velocity is constant.
Also, if the frictional force is more than the applied force, it means the object will not move because the force of resistance is more than the applied force.Finally, if the frictional force is equal to the applied force, then the box will move with a constant motion.
Thus, in conclusion the friction is equal to the applied force.
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calculate the limiting force of friction for a 15 kg object on a surface with a coefficient of dynamic friction of .43.
Answers
The limiting force of friction for a 15 kg object on a surface with a coefficient of dynamic friction of .43 is 63.21 N.
Calculation:-
Normal force = 15 × 9.8 N
= 147 N
Limiting friction is = μN
= 0.43 × 147
= 63.21 N
Friction is the force resisting the relative motion of sturdy surfaces, fluid layers, and material factors sliding towards each specific. There are numerous styles of friction: Dry friction is a strain that opposes the relative lateral movement of two stable surfaces in touch.
Friction is the shape of contact pressure. It exists between the surfaces that are in touch. The frictional force relies upon the character of the floor in contact. The rougher the ground, the extra friction is involved. The frictional force is proportional to the urgent pressure, which is the burden of the body. it's miles impartial of the place of the touch.
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a convex mirror has a focal length of -12 cm. a lightbulb with diameter of 6 cm is placed 60 cm from the mirror
Answers
Here the focal length of the convex mirror as -12 cm, a light bulb with a diameter of 6 cm is placed 60 cm from the mirror. Hence, the image distance is -60 cm, and the image height is 6 cm.
To find the image distance and height, we will use the mirror formula for a convex mirror, which is given as follows;1/f = 1/v + 1/u. Where, f = focal length of the convex mirror;
v = the image distance;
u = the object distance
If the object distance is positive, it means the object is placed in front of the mirror, and if it is negative, it means the object is placed behind the mirror. The focal length of a convex mirror is always negative, as given above.
Therefore, u = -60 cm and f = -12 cm.
Plugging in the values in the mirror formula, we get:1/-12 = 1/v + 1/-60=> -5/60 = 1/v - (1/60) => -5/60 = (60 - v)/60=> v = -300/5= -60 cm
The image distance is -60 cm, which means the image is virtual and erect. The negative sign indicates that the image is formed behind the mirror.
To find the image height, we use the magnification formula:
m = -v/u= -(-60)/(-60)= 1
The magnification is 1, which means the image is of the same size as the object. The height of the light bulb is 6 cm, so the height of the image will also be 6 cm.
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Order these space notes in the bass clef from highest space to lowest space?
1) E
2)C
3)A
4)G
Answers
Answer:
g e c a
Explanation:
Hope I helped!
The air in a tire pump has a volume of 1.50L at a temperature of 5◦C. If the temperature is increased to 30◦C and the pressure remains constant, what is the new volume? Hurry Please!
Answers
Answer:
1.6L
Explanation:
Given parameters:
Initial volume = 1.5L
Initial temperature = 5°C = 5 + 273 = 278K
Final temperature = 30°C = 30 + 273 = 303K
Unknown:
Final or new volume = ?
Solution;
Since pressure is constant, we simply apply Charles's law to solve this problem.
Using the expression below,
\(\frac{V_{1} }{T_{1} }\) = \(\frac{V_{2} }{T_{2} }\)
V and T are volume and temperature
1 and 2 are initial and final states
Insert the parameters and solve;
\(\frac{1.5}{278}\) = \(\frac{V_{2} }{303}\)
278V₂ = 1.5 x 303
V₂ = 1.6L
Science ck-12 green house effect
Answers
Answer:
what about it i need more info
Explanation:
What angle is necessary to keep a 10 kg box motionless if the coefficient of static friction between the box and the ramp is 0.55?
a.33.4°
b.28.8°
c.56.6°
d.45.0°
Answers
The angle necessary to keep a 10 kg box motionless, given a coefficient of static friction of 0.55 between the box and the ramp, is 33.4°, which corresponds to Option A.
To determine the angle, we can use the relationship between the coefficient of static friction, the angle of the incline, and the gravitational force acting on the box. The maximum static friction force can be calculated using the formula:
Friction force = coefficient of static friction * Normal force
The Normal force can be found by decomposing the gravitational force acting on the box into components parallel and perpendicular to the incline. The perpendicular component (Normal force) is equal to the weight of the box (mass * gravitational acceleration).
Since the box is motionless, the friction force must be equal to the component of the gravitational force acting parallel to the incline:
Friction force = Component of weight parallel to incline
By substituting the given values and solving for the angle, we find:
coefficient of static friction = tan(angle)
angle = arctan(coefficient of static friction)
angle = arctan(0.55) ≈ 33.4°
Therefore, the correct answer is Option A, 33.4°.
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Directions: Complete the concept map using the terms listed below.
pushes and pulls
Newton's Laws
inertia
acceleration
unbalanced
mass
pairs
gravity
1.
and
forces
such as
when they are
7.
2.
they
generate
which is a
force of
attraction
motion
can be measure by
between two
objects with
3.
4.
8.
describes
which is a
measure
of
relates force,
mass and
describes
forces in
5.
6.
Answers
Answer:
this is confusing
Explanation:
What are some examples of EM waves
Answers
Answer:
MicrowavesHeatwaves Ultraviolent waves WiFi signal4G/5G signalsExplanation:
All of these are examples of electromagnetic radiation or energy.
The spectra comprises most energy and matter we can account for so far.
The main distinction between EM waves and for example longitudinal acoustic waves is the medium of energy transfer (and rules). EMWs are subject to the idiosyncratic effects of Quantum Mechanics while other 'macro' waves whose medium relies on gross material, are apparently not.
Which of the following particle is responsible for the transfer of static charge?
A. Neutron
B. Electron
C. Proton
D. A & C
Answers
Answer:
B. Electron
Static charge is produced by electron transfer.
Explanation:
Answer A: The neutron does not possess a charge and is said to be neutral.
Answer D: Protons and neutrons never move from object to object.
Only negative charges can move freely from one object to another. The energy that comes from these charges particles is called electrical energy.
....
Hope this answer can help you.
The answer is an electron.
I need help because I don't get it
Answers
assuming that the air is still, how long would it take a large raindrop (5 mm diameter) to reach the ground if it fell from a cloud base at 3000 meters? how long would a typical raindrop (2 mm) take to fall to the ground from the same cloud? how long if it were a drizzle drop (0.5 mm)?
Answers
A typical raindrop would take 500 seconds
How long would a typical raindrop (2 mm) take to fall to the ground from the same cloud?
To calculate the time it takes for a raindrop to fall to the ground, we can use the formula: time = distance / speed. The terminal velocities for different raindrop sizes are approximately:
- Large raindrop (5 mm): 9 m/s
- Typical raindrop (2 mm): 6 m/s
- Drizzle drop (0.5 mm): 2 m/s
Assuming the cloud base is at 3000 meters, we can calculate the time for each type of raindrop as follows:
Large raindrop (5 mm): time = 3000m / 9 m/s = 333.33 seconds
Typical raindrop (2 mm): time = 3000m / 6 m/s = 500 seconds
Drizzle drop (0.5 mm): time = 3000m / 2 m/s = 1500 seconds
So, a large raindrop would take about 333.33 seconds to reach the ground,A normal raindrop takes 500 seconds to form., and a drizzle drop would take 1500 seconds to fall from the cloud base at 3000 meters to the ground when the air is still.
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Question 27
What would be easier for someone with a negative Weber slope for weight detection?
A. Telling the difference between 1 pound and 2 pounds
B. Telling the difference between 10 pound and 11 pounds
C. Telling the difference between 20 pound and 21 pounds
D. No weight discriminations could be made
Answers
For someone with a negative Weber slope for weight detection, it would be easier to tell the difference between 20 pounds and 21 pounds.
The Weber's Law states that the just noticeable difference (JND) between two stimuli is proportional to the magnitude of the stimuli. A negative Weber slope indicates that the JND decreases as the magnitude of the stimuli increases. In this case, as the weight increases, the person's ability to discriminate between the weights becomes easier.
Therefore, option C, telling the difference between 20 pounds and 21 pounds, would be easier for someone with a negative Weber slope for weight detection.
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Which symbol represents a type of radiation that has the same mass as an
electron and a positive charge?
Answers
Answer:
the answer is D
as an electron is composed of positive charges
Answer:
D.+1°e symbol represents a type of radiation that has the same mass as an electron and a positive charge.
Explanation:
it is 100% correct!
what is the velocity of a plane that traveld 3000 miles from new york to california in 5.0 hours
Answers
Does this graph represent an endothermic or an exothermic reaction
Answers
Answer:
the graph represents an endothermic reaction
Explanation:
remember:
endo- inside
exo- out
what will you do to determine the right quantity of the materials needed
Answers
Answer:
Analyse the material required to build the instrument gauge and then keep a buffer stock of 1%
Explanation:
The right quantity to buy differs for every material. There are some materials which are very expensive and are very less used than buffer stock for such materials cannot be purchased. Usually the buffer is to be on the safe side from the shortage of the material. The buffer stock kept is around 1 to 2% for metal materials.
4) A force of 500 N acts on an area of 0.05m2. Find the pressure in Pascal.
Answers
Answer:
pressure = force ie 500 N divided by area ie 0.05m².
p=f by a
p= 500n divided by 0.05 m²
p= 10,000 pascal
Con respecto a las magnitudes longitud del péndulo y periodo, ¿por qué la línea de mejor ajuste de los puntos-dato de la distribución indica que sí hay relación entre ambas magnitudes?
Answers
Answer:
Por que existe una relación entre dichas magnitudes.
Para un pendulo perfecto de largo L, sabemos que el periodo esta definido como:
\(T = 2*\pi *\sqrt{L/g}\)
Donde:
pi = 3.14
L = largo del péndulo
g = aceleración gravitatoria = 9.8 m/s^2
Entonces podemos ver que el periodo es proporcional a la raíz cuadrada de la longitud del péndulo.
Esto es por lo que al analizar datos de un experimento se debería observar una relación entre esas magnitudes.
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 3. 20×10−19 J. In what direction and through what potential difference Vb−Va does the particle move?a. The particle moves to the left through a potential difference of Vb-Va = 0. 500 V. B. The particle moves to the left through a potential difference of Vb-Va = -0. 500 Vc. The particle moves to the right through a potential difference of Vb-Va = 0. 500 V. D. The particle moves to the right through a potential difference of Vb-Va = -0. 500 V. E. The particle moves to the left through a potential difference of Vb-Va = 5. 00 V. F. The particle moves to the right through a potential difference of Vb-Va = -5. 00 V
Answers
The particle moves to the right through a potential difference of Vb-Va = 3.20×10^-19 J / q.
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 3.20×10^-19 J. To determine the direction and potential difference through which the particle moves, we can use the relationship between electric potential difference (V) and electric force (F) on a charged particle.
The electric potential difference between two points is defined as the work done per unit charge to move a particle from one point to the other. The work done on a charged particle by an electric force is given by the equation
W = Fdcos(theta)
where F is the electric force, d is the distance moved by the particle, and theta is the angle between the direction of the force and the direction of the displacement.
In this case, since the particle is only acted upon by the electric force and moves along the x-axis, the angle between the direction of the force and the direction of the displacement is 0, and the work done by the electric force is given by W = Fdx.
Therefore, the potential difference between point a and point b is given by Vb - Va = W/q, where q is the charge of the particle.
Given that the particle's kinetic energy increases by 3.20×10^-19 J and the work-energy principle states that work done on a particle is equal to the change in kinetic energy, we can say that the work done on the particle is equal to 3.20×10^-19 J.
Now, the direction of the force can be determined by the sign of the potential difference, since the electric force is given by
F = -q(dV/dx).
Given that the potential difference is positive, the electric force is negative, meaning that the force is directed opposite to the direction of motion of the particle, therefore the direction of motion is to the right.
Therefore, the particle moves to the right through a potential difference of Vb-Va = 3.20×10^-19 J / q.
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