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Morgarella [4.7K]
3 years ago
12

Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.

Physics
2 answers:
masya89 [10]3 years ago
5 0
The velocity would be the same for all ramps.
timama [110]3 years ago
4 0

Answer:

The ball would have greatest speed at the bottom of ramp C

Explanation:

From the given ramps, C would generate the highest potential energy especially at the highest point that the ball starts to move down the ramp. Potential energy is the energy possessed by an object moving with a uniform velocit or at rest. Since potential energy increases with height, the potential energy gained by the ball on ramp C would cause an increase in its momentum as it glides down to the bottom of the ramp.

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4 0
2 years ago
A freight car moves along a friction less level railroad track at constant speed. The car is open on top. A large load of coal i
maxonik [38]

Answer:

The velocity of the freight car decreases.

Explanation:

This question is answered by the conservation of momentum principle.

When the freight car is moving at a certain speed, it has a constant momentum.

We will call this M1.

The equation for M1 will be:

M1 = Mass * Speed

Now when the coal is dumped into the freight car, the Mass increases.

Since conservation of momentum states that the momentum will remain the same. We have:

M1 = (Mass of freight + Mass of coal) * Speed

Since M1 is constant, if the mass increases, the speed had to decrease to keep the equation true.

8 0
3 years ago
A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak
GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg          

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.  

From the attached figure, the x and y component of forces is given by :

T_{1x} =-T_1 cos (20)

T_{2x} = T_2 cos (40)

mg_x = 0

T_{1y} = T_1 sin (20)

T_{2y} = T_2 sin (40)

mg_y= -mg

Let R_x and R_y is the resultant in x and y direction.

R_x=-T_1 cos (20)+T_2 cos (40)+0

R_y=T_1 sin(20)+T_2 sin(40)-mg

As the system is balanced the net force acting on it is 0. So,

-T_1 cos (20)+T_2 cos (40)+0=0.............(1)

T_1 sin(20)+T_2 sin(40)-100\times 9.8=0..................(2)

On solving equation (1) and (2) we get:  

T_1=866.86\ N (tension on the left rope)

T_2=1063.36\ N (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.                            

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3 years ago
According to Newton, which of these will determine if and how an object will move?
mr Goodwill [35]

Answer:

3

Explanation:

5 0
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