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Gre4nikov [31]
4 years ago
5

A rubber ball rolls across a level surface.The ball will eventually stop rolling due to

Physics
2 answers:
kaheart [24]4 years ago
8 0
The answer would be friction
Illusion [34]4 years ago
7 0
C. Friction is the answer
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A 1.5 kg rock is dropped from a height of 2.0 meters onto a spring that
Maslowich

Answer:

29.4 J

Explanation:

Before the drop, the system has only the gravitational potential energy, and this energy us given by mass×gravity×height:

1.5•9.8•2 = 29.4 J

7 0
3 years ago
A helicopter's speed increases
sergij07 [2.7K]

Answer:

the helis speed increased its speed by 35 m/s. Unless you are asking me what was the speed per second. if you were asking me for the speed per second then that would be a different answer

Explanation:

7 0
3 years ago
Susie walks 3 blocks north to the local CVS store, then 4 blocks east to her grandmother’s house. She then walks 2 blocks west a
Slav-nsk [51]

Answer:

Suzie is 3 blocks north of where she started

Explanation:

Displacement is the minimum distance between the initial and final point of motion.

Here, Suzie first walks 3 blocks north. From there she walks 4 blocks east. Then 2 blocks to the east then 2 blocks north and then 2 blocks east. She covered 4 blocks east toward west. This is the same distance she covered traveling east. But she is 2 blocks north. From there she traveled a block south to the pizzeria and another block to her friends house. She covered the two block she had traveled north.

Hence, Suzie is 3 blocks north of where she started.

7 0
3 years ago
a football is thrown upward at a 31° angle to the horizontal.the acceleration of gravity is 9.8m/s^2. To throw the ball a distan
mr Goodwill [35]

The ball's vertical position y in the air at time t is

y=v_0\sin31^\circ\,t-\dfrac g2t^2

The ball is at its original height when y=0, which happens at

v_0\sin31^\circ\,t-\dfrac g2t^2=\dfrac t2\left(2v_0\sin31^\circ-gt)=0

\implies t=0\text{ and }t=\dfrac{2v_0\sin31^\circ}g

Meanwhile, the ball's horizontal position x at time t is

x=v_0\cos31^circ\,t

So when the ball reaches its original height a second time, the ball will have traveled a horizontal distance of

x=\dfrac{2{v_0}^2\sin31^\circ\cos31^\circ}g=\dfrac{{v_0}^2\sin(2\cdot31^\circ)}g

(which you might recognize as the formula for the range of a projectile)

To reach a distance of x=77\,\rm m, the initial speed v_0 would be

77\,\mathrm m=\dfrac{{v_0}^2\sin62^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\implies v_0=29\dfrac{\rm m}{\rm s}

7 0
3 years ago
The buoyant force acts in all directions. Please select the best answer from the choices provided T F
tester [92]
Where are the answer from the choices ?
5 0
3 years ago
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