What is the mass of an object that is moving, №18009889, 17.07.2022 09:49

Chemistry

What is the mass of an object that is moving at 8.0 m/s and has 70.0 J of kinetic energy?

Step-by-step answer

P Answered by Master

Explanation:

$ke = \frac{1}{2} \times m \times {v}^{2 } \\ m = \frac{2 \times ke}{ {v}^{2} } \\ m = \frac{2 \times 70.0}{ {8}^{2} } \\ m = \frac{140}{64} = 2.1875$

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Physics

30. Easy Guided Online Tutorial One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 m/s. The masses of the two objects are 3.0 and 8.0 kg. Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object

Step-by-step answer

P Answered by Specialist

$18.18\ \text{m/s}$

$6.82\ \text{m/s}$

Explanation:

m_1 = Mass of large object = 8 kg

m_2 = Mass of smaller object = 3 kg

When large mass is moving

u_1 = 25 m/s

u_2 = 0

For completely inelastic collision we have the relation

$m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 25+3\times 0}{8+3}\\\Rightarrow v=18.18\ \text{m/s}$

Speed of the combined mass when the larger object is moving is $18.18\ \text{m/s}$

When smaller mass is moving

u_1 = 0

u_2 = 25 m/s

$v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 0+3\times 25}{8+3}\\\Rightarrow v=6.82\ \text{m/s}$

Speed of the combined mass when the smaller object is moving is $6.82\ \text{m/s}$

Physics

Question 1 According to the Law of Conservation of Energy, why does the first hill on a roller coaster always have to be the tallest of all the other hills? Question 2 A single marble is placed at the top of a hill. The initial potential energy of the marble is 550 J. The marble slides down the hill. You take a picture of the marble at some point as it is down the hill. Without knowing the speed, the mass or the position of the marble, which of the following could be the only possible correct energies of the marble at this point? Assume no friction

Step-by-step answer

P Answered by Master

4

Dang that’s a lot wait i will answer in comments

Physics

Question 1 According to the Law of Conservation of Energy, why does the first hill on a roller coaster always have to be the tallest of all the other hills? Question 2 A single marble is placed at the top of a hill. The initial potential energy of the marble is 550 J. The marble slides down the hill. You take a picture of the marble at some point as it is down the hill. Without knowing the speed, the mass or the position of the marble, which of the following could be the only possible correct energies of the marble at this point? Assume no friction

Step-by-step answer

P Answered by Master

4

Dang that’s a lot wait i will answer in comments

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

Physics 80 joe can pitch a baseball a distance of 48 meters in 1.5 seconds. how fast is his pitch? a) 3.1 m/s b) 32 m/s c) 48 m/s d) 72 m/s 2) if c moves to position 5 (60 meters), what is its average velocity during these 5 seconds? a) 10 m/s b) 11 m/s c) 12 m/s d) 13 m/s 3) consider the motion diagram. it illustrates a car s velocity (v) and acceleration (a). the best description of a car’s velocity in this diagram is: a) a car’s velocity remained the same at 10 m/s. b) a car’s velocity decreased from 10 m/s to 5 m/s. c) a car increased its velocity

Step-by-step answer

P Answered by Specialist

1. B) 32 m/s

The speed of the baseball is given by

$v=\frac{d}{t}$

where

d = 48 m is the distance travelled

t = 1.5 s is the time taken

Substituting,

$v=\frac{48}{1.5}=32 m/s$

2. C) 12 m/s

The average velocity is given by

$v=\frac{d}{t}$

where

d = 60 - 0 = 60 m is the displacement

t = 5 s is the time interval

Substituting,

$v=\frac{60}{5}=12 m/s$

3. Missing diagram

4. D) velocity

In fact, velocity is defined as the ratio between the displacement ( $\Delta s$ ) and the time interval ( $\Delta t$ ) needed to achieve that change in position:

$v=\frac{\Delta s}{\Delta t}$

Similarly, acceleration is defined as the ratio between the change in velocity ( $\Delta v$ ) and the time interval ( $\Delta t$ ):

$a=\frac{\Delta v}{\Delta t}$

Comparing the two equations, we see that displacement is to velocity as velocity is to acceleration.

5. C) 5/1 cm/year

We know that the ridge moves 25 cm in 5 years, so we have:

- Displacement (d): d = 25 cm

- Time interval (t): t = 5 y

Using the equation for the velocity:

$v=\frac{d}{t}=\frac{25 cm}{5 y}=5 cm/y$

6. D

The missing graph is attached. Each graph represents a distance (on the y-axis, measured in metres) versus a time (on the x-axis, measured in seconds). Therefore, the speed in each graph can be simply calculated from the slope of the curve, since speed is the ratio between distance and time:

$v=\frac{\Delta y}{\Delta x}$

For each graph:

A. $\frac{\Delta y}{\Delta x}=\frac{8.6-2}{10}\sim 0.66$

B. $\frac{\Delta y}{\Delta x}=\frac{5.7-0}{10}\sim 0.57$

C. $\frac{\Delta y}{\Delta x}=\frac{9.6-3}{10}\sim 0.66$

D. $\frac{\Delta y}{\Delta x}=\frac{7.5-0}{5}\sim 1.5$

So we can say that the correct graph must be graph D, the closest to 1.57.

7. B) The car has come to a stop and has zero velocity

The missing graph is attached.

The graph shows the position of the car versus time. From the graph, we can see that in segment B, the position of the car does not change: this means that its velocity is zero, since the velocity is defined as the ratio between the change in position (displacement) and the time interval:

$v=\frac{\Delta s}{\Delta t}$

And since the displacement is zero, the velocity is also zero.

8. C) from 4.5 to 5.0 seconds

Missing graph is attached.

The graph represents the distance covered versus the time taken: therefore, the speed of the ball can be evaluated from the slope of the curve.

The only region in which the slope of the curve is zero is between 4.5 seconds and 5.0 seconds: therefore, this is the region where the soccer ball is not moving.

9. D) diagonal line with varying slope, from 3 to 4.

The table of data is:

Time (s) 0 1 2 3 4

Distance (m) 0 3 6 12 16

We want to know how the distance/time graph would like like. We have:

- At t = 1, the distance is d=3 , so the slope is $\frac{d}{t}=\frac{3}{1}=3$

- Similarly, at t = 2: $\frac{d}{t}=\frac{6}{2}=3$

- Instead, at t =3, the slope is $\frac{d}{t}=\frac{12}{3}=4$

- Similarly, at t=4, $\frac{d}{t}=\frac{16}{4}=4$

So the correct answer is D.

10. D) To move in a circle requires an acceleration and therefore a net force. This force is supplied by the slingshot.

For an object in circular motion, the velocity constantly changes (because the direction changes): this means that an acceleration is required (towards the centre of the circle), and therefore a force must be exerted (also towards the centre of the circle) to keep the object in the circular path.

As soon as this force is removed, the object is "free" to leave the circular trajectory and continue its motion following a straight path at constant velocity, according to Newton's first law.

11. B) The acceleration will become 1/4 as much.

Re-arranging Newton's second law formula,

$F=ma \rightarrow a = \frac{F}{m}$ (1)

The force exerted on the satellite is actually the force of gravity:

$F=\frac{Gm M}{d^2}$ (2)

Substituting (2) into (1),

$a=\frac{F}{m}=\frac{GM}{d^2}$

We see that there is no dependance on the mass of the satellite (m), but only on the distance. Since in this problem the distance is doubled (d'=2d), the new acceleration will be:

$a'=\frac{GM}{(2d)^2}=\frac{1}{4}(\frac{GM}{d^2})=\frac{a}{4}$

12. C) divide: distance : velocity

Velocity is distance divided by time:

$v=\frac{d}{t}$

Multiplying by t on both sides,

$v\cdot t = \frac{d}{t} \cdot t \rightarrow vt = d$

and dividing by v on both sides,

$\frac{vt}{v}=\frac{d}{v} \rightarrow t = \frac{d}{v}$

13. A) -8 km/h

Let's call:

v_k = +5 km/h the velocity of the kite

v_p = -3 km/h your velocity (opposite direction)

In your frame of reference, the velocity of the kite will be:

v'_k = v_p - v_k = -3 -(+5) = -8 km/h

Physics

Physics 80 joe can pitch a baseball a distance of 48 meters in 1.5 seconds. how fast is his pitch? a) 3.1 m/s b) 32 m/s c) 48 m/s d) 72 m/s 2) if c moves to position 5 (60 meters), what is its average velocity during these 5 seconds? a) 10 m/s b) 11 m/s c) 12 m/s d) 13 m/s 3) consider the motion diagram. it illustrates a car s velocity (v) and acceleration (a). the best description of a car’s velocity in this diagram is: a) a car’s velocity remained the same at 10 m/s. b) a car’s velocity decreased from 10 m/s to 5 m/s. c) a car increased its velocity

Step-by-step answer

P Answered by Specialist

1. B) 32 m/s

The speed of the baseball is given by

$v=\frac{d}{t}$

where

d = 48 m is the distance travelled

t = 1.5 s is the time taken

Substituting,

$v=\frac{48}{1.5}=32 m/s$

2. C) 12 m/s

The average velocity is given by

$v=\frac{d}{t}$

where

d = 60 - 0 = 60 m is the displacement

t = 5 s is the time interval

Substituting,

$v=\frac{60}{5}=12 m/s$

3. Missing diagram

4. D) velocity

In fact, velocity is defined as the ratio between the displacement ( $\Delta s$ ) and the time interval ( $\Delta t$ ) needed to achieve that change in position:

$v=\frac{\Delta s}{\Delta t}$

Similarly, acceleration is defined as the ratio between the change in velocity ( $\Delta v$ ) and the time interval ( $\Delta t$ ):

$a=\frac{\Delta v}{\Delta t}$

Comparing the two equations, we see that displacement is to velocity as velocity is to acceleration.

5. C) 5/1 cm/year

We know that the ridge moves 25 cm in 5 years, so we have:

- Displacement (d): d = 25 cm

- Time interval (t): t = 5 y

Using the equation for the velocity:

$v=\frac{d}{t}=\frac{25 cm}{5 y}=5 cm/y$

6. D

The missing graph is attached. Each graph represents a distance (on the y-axis, measured in metres) versus a time (on the x-axis, measured in seconds). Therefore, the speed in each graph can be simply calculated from the slope of the curve, since speed is the ratio between distance and time:

$v=\frac{\Delta y}{\Delta x}$

For each graph:

A. $\frac{\Delta y}{\Delta x}=\frac{8.6-2}{10}\sim 0.66$

B. $\frac{\Delta y}{\Delta x}=\frac{5.7-0}{10}\sim 0.57$

C. $\frac{\Delta y}{\Delta x}=\frac{9.6-3}{10}\sim 0.66$

D. $\frac{\Delta y}{\Delta x}=\frac{7.5-0}{5}\sim 1.5$

So we can say that the correct graph must be graph D, the closest to 1.57.

7. B) The car has come to a stop and has zero velocity

The missing graph is attached.

The graph shows the position of the car versus time. From the graph, we can see that in segment B, the position of the car does not change: this means that its velocity is zero, since the velocity is defined as the ratio between the change in position (displacement) and the time interval:

$v=\frac{\Delta s}{\Delta t}$

And since the displacement is zero, the velocity is also zero.

8. C) from 4.5 to 5.0 seconds

Missing graph is attached.

The graph represents the distance covered versus the time taken: therefore, the speed of the ball can be evaluated from the slope of the curve.

The only region in which the slope of the curve is zero is between 4.5 seconds and 5.0 seconds: therefore, this is the region where the soccer ball is not moving.

9. D) diagonal line with varying slope, from 3 to 4.

The table of data is:

Time (s) 0 1 2 3 4

Distance (m) 0 3 6 12 16

We want to know how the distance/time graph would like like. We have:

- At t = 1, the distance is d=3 , so the slope is $\frac{d}{t}=\frac{3}{1}=3$

- Similarly, at t = 2: $\frac{d}{t}=\frac{6}{2}=3$

- Instead, at t =3, the slope is $\frac{d}{t}=\frac{12}{3}=4$

- Similarly, at t=4, $\frac{d}{t}=\frac{16}{4}=4$

So the correct answer is D.

10. D) To move in a circle requires an acceleration and therefore a net force. This force is supplied by the slingshot.

For an object in circular motion, the velocity constantly changes (because the direction changes): this means that an acceleration is required (towards the centre of the circle), and therefore a force must be exerted (also towards the centre of the circle) to keep the object in the circular path.

As soon as this force is removed, the object is "free" to leave the circular trajectory and continue its motion following a straight path at constant velocity, according to Newton's first law.

11. B) The acceleration will become 1/4 as much.

Re-arranging Newton's second law formula,

$F=ma \rightarrow a = \frac{F}{m}$ (1)

The force exerted on the satellite is actually the force of gravity:

$F=\frac{Gm M}{d^2}$ (2)

Substituting (2) into (1),

$a=\frac{F}{m}=\frac{GM}{d^2}$

We see that there is no dependance on the mass of the satellite (m), but only on the distance. Since in this problem the distance is doubled (d'=2d), the new acceleration will be:

$a'=\frac{GM}{(2d)^2}=\frac{1}{4}(\frac{GM}{d^2})=\frac{a}{4}$

12. C) divide: distance : velocity

Velocity is distance divided by time:

$v=\frac{d}{t}$

Multiplying by t on both sides,

$v\cdot t = \frac{d}{t} \cdot t \rightarrow vt = d$

and dividing by v on both sides,

$\frac{vt}{v}=\frac{d}{v} \rightarrow t = \frac{d}{v}$

13. A) -8 km/h

Let's call:

v_k = +5 km/h the velocity of the kite

v_p = -3 km/h your velocity (opposite direction)

In your frame of reference, the velocity of the kite will be:

v'_k = v_p - v_k = -3 -(+5) = -8 km/h

Physics

Which of the following objects in motion has the highest inertia? A.) A 5.0 kg mass moving at 2.0 m/s B.) A 8.0 kg mass moving at 9.0 m/s C.) A 6.0 kg mass moving at 10. m/s D.) A 15 kg mass moving at 2.0 m/s

Step-by-step answer

P Answered by PhD

D.) A 15 kg mass moving at 2.0 m/s

Explanation:

The inertia of an object is most directly related to its mass and not the magnitude of the velocity.

So, the higher the mass of an object, the more its inertia.

According to Newton's first law of motion "an object will remain in its state of rest or of constant motion unless it is acted upon by an external force".

Inertia is the tendency of a body to remain in a state of rest either by virtue of its static position of constant motion. The larger the mass, the more this tendency. An object with a large mass will take more force to bring it from rest or change its motion.

Physics

Which of the following objects in motion has the highest inertia? A.) A 5.0 kg mass moving at 2.0 m/s B.) A 8.0 kg mass moving at 9.0 m/s C.) A 6.0 kg mass moving at 10. m/s D.) A 15 kg mass moving at 2.0 m/s

Step-by-step answer

P Answered by PhD

D.) A 15 kg mass moving at 2.0 m/s

Explanation:

The inertia of an object is most directly related to its mass and not the magnitude of the velocity.

So, the higher the mass of an object, the more its inertia.

According to Newton's first law of motion "an object will remain in its state of rest or of constant motion unless it is acted upon by an external force".

Inertia is the tendency of a body to remain in a state of rest either by virtue of its static position of constant motion. The larger the mass, the more this tendency. An object with a large mass will take more force to bring it from rest or change its motion.

What is the mass of an object that is moving at 8.0 m/s and has 70.0 J of kinetic energy?

Step-by-step answer

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