Physics: 4. A 40 N force is applied to a 10 kg block, but the object only accelerates at 2 m/s2. Explain why this happens using a free-body diagram.

4. A 40 N force is applied to a 10 kg block,, №14266766, 29.09.2020 09:08

Physics

14. a block rests on a frictionless table on earth. after a 20-n horizontal force is applied to the block, it accelerates at 3.9 m/s2. then the block and table are carried to the moon, where the acceleration due to gravity is 1.62 m/s2. a horizontal force of 10 n is then applied to the block. what is the acceleration? a. 1.8 m/s2 b. 2.5 m/s2 c. 2.1 m/s2 d. 2.3 m/s2 e. 2.0 m/s2 15. workers are trying to free an suv stuck in the mud. to move the vehicle, they use three ropes held parallel to the ground, producing the force vectors shown in the figure.

Step-by-step answer

P Answered by PhD

5

14. E 2.0 m/s^2

Initially, a 20 N force is applied to the block, so it has an acceleration of 3.9 m/s^2. According to Newton's law, the mass of the block is:

$m=\frac{F}{a}=\frac{20 N}{3.9 m/s^2}=5.13 kg$

In the second situation, a force of 10 N is applied to the block. Since the mass is still the same, the acceleration now is:

$a=\frac{F}{m}=\frac{10 N}{5.13 kg}=1.95 m/s^2$

So, approximately 2.0 m/s^2.

15. C. 866 N at 78.1° counterclockwise to the x-axis

Resultant along the x- and y-axis:

$R_x = (985 N)(cos 31^{\circ})-(788 N)(sin 32^{\circ})-(411 N)(cos 53^{\circ})=179.4 N$

$R_y = (985 N)(sin 31^{\circ})+(788 N)(cos 32^{\circ})-(411 N)(sin 53^{\circ})=847.3 N$

Magnitude and direction of the resultant:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(179.4 N)^2+(847.3 N)^2}=866.0 N$

$\theta=arctan(\frac{R_y}{R_x})=\arctan(\frac{847.3 N}{179.4 N})=78.1^{\circ}$

16. D. 1720 N

Since the sphere is suspended, it is in equilibrium, therefore the tension in the chain is equal to the weight of the sphere attached to it, therefore:

T=mg=(175 kg)(9.81 m/s^2)=1720 N

17. C. A

This arrangement generates the largest tension in the chain, because in all other arrangements the weight of the object is split between the two chains, while in this case all the weight is hold by one chain, therefore the tension in this case is larger.

18. Straight path

The gravity "holds" the planets keeping them in a circular orbit. If we remove gravity, the planets would continue in a straight path with constant speed, because now there are no more forces acting on it, so by inertia they will continue their uniform motion with constant speed.

19. B. 1.6 × 104 N

First we can find the deceleration of the car by using the SUVAT equation:

v^2 -u^2 =2aS

where v=0 m/s, u=2 m/s, and S=15 cm=0.15 m. Re-arranging, we have

$a=\frac{-u^2}{2S}=\frac{-(2 m/s)^2}{2(0.15 m)}=-13.3 m/s^2$

And now we can calculate the average force exerted on the car, by using Newton's second law:

$F=ma=(1200 kg)(-13.3 m/s^2)=-15960 N=-1.6 \cdot 10^4 N$

(the negative sign means that the force's direction is opposite to the motion of the car)

20. A. magnetism

Magnetism is part of the electromagnetic force, which is one of the fundamental forces which act also through empty space. All the other forces need some object in order to act.

21. E. 45 N

The magnitude of the force in link A is equal to the weight of the rod plus the weight of the lower block, therefore:

W=(m_1 + m_2)g=(0.6 kg+4.0 kg)(9.8 m/s^2)=45 N

22. C. on Earth at sea level

The weight of the bowling ball is given by: W=mg , where m is the mass of the ball and g is the acceleration due to gravity. The value of g increases when moving from the Earth's center to the Earth's surface, then decreases when moving far from the surface, so the point where g is greatest is at sea level, where it is 9.81 m/s^2. On the surface of the Moon, g is much smaller (about 1/6 of the value on Earth).

23. A. 0.5 m/s2

The acceleration of the block is given by Newton's second law:

$a=\frac{F}{m}=\frac{20 N}{40 kg}=0.5 m/s^2$

24. D. Both forces are equal in magnitude but opposite in direction.

According to Newton's third law: if an object A exerts a force on an object B, then object B exerts a force equal and opposite on object B. In this case, objects A and B are the bat and the baseball, therefore the two forces are equal in magnitude and opposite in direction.

25. B. 2940 N

The mass of the boulder is equal to its weight divided by the acceleration of gravity (9.81 m/s^2):

$m=\frac{W}{g}=\frac{2400 N}{9.81 m/s^2}=245 kg$

So now we can calculate the force needed to accelerate the boulder to 12.0 m/s^2:

F=ma=(245 kg)(12.0 m/s^2)=2940 N

26. D. 32.2 N

The weight of the object on Mercury is given by:

W=mg=(8.69 kg)(3.71 m/s^2)=32.2 N

Physics

14. a block rests on a frictionless table on earth. after a 20-n horizontal force is applied to the block, it accelerates at 3.9 m/s2. then the block and table are carried to the moon, where the acceleration due to gravity is 1.62 m/s2. a horizontal force of 10 n is then applied to the block. what is the acceleration? a. 1.8 m/s2 b. 2.5 m/s2 c. 2.1 m/s2 d. 2.3 m/s2 e. 2.0 m/s2 15. workers are trying to free an suv stuck in the mud. to move the vehicle, they use three ropes held parallel to the ground, producing the force vectors shown in the figure.

Step-by-step answer

P Answered by PhD

5

14. E 2.0 m/s^2

Initially, a 20 N force is applied to the block, so it has an acceleration of 3.9 m/s^2. According to Newton's law, the mass of the block is:

$m=\frac{F}{a}=\frac{20 N}{3.9 m/s^2}=5.13 kg$

In the second situation, a force of 10 N is applied to the block. Since the mass is still the same, the acceleration now is:

$a=\frac{F}{m}=\frac{10 N}{5.13 kg}=1.95 m/s^2$

So, approximately 2.0 m/s^2.

15. C. 866 N at 78.1° counterclockwise to the x-axis

Resultant along the x- and y-axis:

$R_x = (985 N)(cos 31^{\circ})-(788 N)(sin 32^{\circ})-(411 N)(cos 53^{\circ})=179.4 N$

$R_y = (985 N)(sin 31^{\circ})+(788 N)(cos 32^{\circ})-(411 N)(sin 53^{\circ})=847.3 N$

Magnitude and direction of the resultant:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(179.4 N)^2+(847.3 N)^2}=866.0 N$

$\theta=arctan(\frac{R_y}{R_x})=\arctan(\frac{847.3 N}{179.4 N})=78.1^{\circ}$

16. D. 1720 N

Since the sphere is suspended, it is in equilibrium, therefore the tension in the chain is equal to the weight of the sphere attached to it, therefore:

T=mg=(175 kg)(9.81 m/s^2)=1720 N

17. C. A

This arrangement generates the largest tension in the chain, because in all other arrangements the weight of the object is split between the two chains, while in this case all the weight is hold by one chain, therefore the tension in this case is larger.

18. Straight path

The gravity "holds" the planets keeping them in a circular orbit. If we remove gravity, the planets would continue in a straight path with constant speed, because now there are no more forces acting on it, so by inertia they will continue their uniform motion with constant speed.

19. B. 1.6 × 104 N

First we can find the deceleration of the car by using the SUVAT equation:

v^2 -u^2 =2aS

where v=0 m/s, u=2 m/s, and S=15 cm=0.15 m. Re-arranging, we have

$a=\frac{-u^2}{2S}=\frac{-(2 m/s)^2}{2(0.15 m)}=-13.3 m/s^2$

And now we can calculate the average force exerted on the car, by using Newton's second law:

$F=ma=(1200 kg)(-13.3 m/s^2)=-15960 N=-1.6 \cdot 10^4 N$

(the negative sign means that the force's direction is opposite to the motion of the car)

20. A. magnetism

Magnetism is part of the electromagnetic force, which is one of the fundamental forces which act also through empty space. All the other forces need some object in order to act.

21. E. 45 N

The magnitude of the force in link A is equal to the weight of the rod plus the weight of the lower block, therefore:

W=(m_1 + m_2)g=(0.6 kg+4.0 kg)(9.8 m/s^2)=45 N

22. C. on Earth at sea level

The weight of the bowling ball is given by: W=mg , where m is the mass of the ball and g is the acceleration due to gravity. The value of g increases when moving from the Earth's center to the Earth's surface, then decreases when moving far from the surface, so the point where g is greatest is at sea level, where it is 9.81 m/s^2. On the surface of the Moon, g is much smaller (about 1/6 of the value on Earth).

23. A. 0.5 m/s2

The acceleration of the block is given by Newton's second law:

$a=\frac{F}{m}=\frac{20 N}{40 kg}=0.5 m/s^2$

24. D. Both forces are equal in magnitude but opposite in direction.

According to Newton's third law: if an object A exerts a force on an object B, then object B exerts a force equal and opposite on object B. In this case, objects A and B are the bat and the baseball, therefore the two forces are equal in magnitude and opposite in direction.

25. B. 2940 N

The mass of the boulder is equal to its weight divided by the acceleration of gravity (9.81 m/s^2):

$m=\frac{W}{g}=\frac{2400 N}{9.81 m/s^2}=245 kg$

So now we can calculate the force needed to accelerate the boulder to 12.0 m/s^2:

F=ma=(245 kg)(12.0 m/s^2)=2940 N

26. D. 32.2 N

The weight of the object on Mercury is given by:

W=mg=(8.69 kg)(3.71 m/s^2)=32.2 N

Physics

The system shown above consists of two identical blocks that are suspended using four cords, each of a different length. Which of the following claims are true about the magnitudes of the tensions in the cords?

Step-by-step answer

P Answered by Specialist

Answer: Option B and C are True.

Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂:
w = T₁/cos 60° -----(1);
w = T₂/cos 30° ----(2);
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°;
T₁ cos 30° = T₂ cos 60°;
T₂/T₁ = cos 30°/cos 60°;
T₂/T₁ =1.73.
Therefore, option a is false since T₂ > T₁.
Option B is true since T₁ cos 30° = T₂ cos 60°.
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
Answer: Option B and C are True.

Explanation:
The weight of the two blocks acts downwards.
Le

Physics

A proton (1.6726 ? 10-27 kg) and a neutron (1.6749 ? 10-27 kg) at rest combine to form a deuteron, the nucleus of deuterium or “heavy hydrogen.” In this process, a gamma ray (high-energy photon) is emitted, and its energy is measured to be 2.2 MeV (2.2 ? 10 6 eV).

Step-by-step answer

P Answered by Master

Answer:

see below.

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

Physics

tomatoes are being thrown horizontally off the top of a building 50.0 meters above the ground at 3.00 m/s. How far will the tomatoes travel horizontally by the time it hits the ground? Use significant figures, and estimate g to be equal to 9.81 m/s^2

Step-by-step answer

P Answered by PhD

Answer:

9.6 meters

Step-by-step explanation:

Time taken by the tomatoes to each the ground

using h = 1/2 g t^2

t^2 = 2h/g = 2 x 50/ 9.8 = 10.2

t = 3.2 sec

horizontal ditance = speed x time = 3 x 3.2 = 9.6 meters

Physics

a screw has a head diameter of 0.812 m and a thread width of 0.318 cm. What is the ideal mechanical advantage?

Step-by-step answer

P Answered by PhD

1

The question specifies the diameter of the screw, therefore the IMA of this screw is 0.812? / 0.318 = 8.02

Physics

A puck moves 2.35 m/s in a -22.0° direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50.0° direction. What was the magnitude of the acceleration?

Step-by-step answer

P Answered by PhD

Answer:
7.25 secs.

Explanation:
First find the distance it takes to stop
s = [v^2-u^2]/2a = 0^2 - 8.7^2/2[-2.4] = 8.7^2/4.8
Next find the time it takes to go that distance , s = ut +[1/2] at^2
8.7^2/4.8 = 8.7t +[1/2] [ -2.4]t^2 , rearrange and
t^2 -[8.7/1.2]+ 8.7^2/[(1.2)(4.8)]=0 complete the square
[t - (8.7/2.4)]^2=0
t = 8.7/2.4 = 3.625 secs
At this stage the deceleration will push the object back in the direction it came from for another 3.625 secs when it will be 8.7 m/s again
Total time , T =2t = 7.25 secs.

Note:
The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity. For example, if x is a variable, then a change in the value of x is often denoted Δx (pronounced delta x). The differential dx represents an infinitely small change in the variable x.

Physics

An electrical kettle supplies 20,000j of energy to heat 0.5kg of water. What is the temperature change? The specific heat capacity of water is 4200j/kg.

Step-by-step answer

P Answered by PhD

The change in temperature is 9.52°CExplanation:Since, the heat supplied by the electric kettle is totally used to increase the temperature of the water.Thus, from the law of conservation of energy can be stated as:Heat Supplied by Electric Kettle = Heat Absorbed by WaterHeat Supplied by Electric Kettle = m C ΔTwhere,Heat Supplied by Electric Kettle = 20,000 JMass of water = m = 0.5 kgSpecific Heat Capacity of Water = C = 4200 J/kg.°CChange in Temperature of Water = ΔTTherefore,20,000 J = (0.5 kg)(4200 J/kg.°C) ΔTΔT = 20,000 J/(2100 J/°C)ΔT = 9.52°C

Physics

Jordan lifts a 100-kilogram barbell from the floor to a height of 2.0 meters in 1.5 seconds. The amount of power he generates is about ____ watts.

Step-by-step answer

P Answered by PhD

1

Weight of barbell (m) = 100 kg
Uplifted to height (h) = 2m
Time taken= 1.5 s
Work done by Jordan = potential energy stored in barbell = mgh
= 100×2×9.8
= 1960J
Power = energy/time
= 1960/1.5
1306.67watts

Physics

Klay Thompson and Draymond Green wanted to play a joke on Steph Curry. Klay threw a ballon horizontally at 50 m/s from 150 m high bridge. How far did the ballow go when it landed?

Step-by-step answer

P Answered by PhD

The horizontal and vertical motions of balloons are independent from each other.
Let vertical component of initial velocity U' horizontal component of initial velocity U"
Time of landing (t) is found with the help of vertical motion.
Since vertical component of initial velocity of balloon is zero(U' = 0)
From equation h = U't + 1/2gt^2
h = 1/2gt^2
t = √(2h/g)
t = √( 2×150/9.8)
t = 5.53 sec
Horizontal velocity = 50m/s
Horizontal range of balloon, R = U"t
= 50× 5.53
= 27.65m
So the balloon will go 27.65 metre away from the bridge

4. A 40 N force is applied to a 10 kg block, but the object only accelerates at 2 m/s2. Explain why this happens using a free-body diagram.

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