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WINSTONCH [101]
3 years ago
8

Jacob is riding his bike down a tall hill. At the top of the hill, he has 2800J of potential energy. At the bottom of the hill,

when he is going the fastest, his kinetic energy measures 2650J. Which statement below is true? Question 5 options: The total amount of energy at the bottom of the hill cannot equal the amount at the top, because some of the energy is gained. The total amount of energy at the bottom is equal to the amount at the top, but some energy was converted into other forms due to friction. The total amount of energy at the bottom of the hill cannot equal the amount at the top, because some of the energy is destroyed. The total amount of energy at the bottom is equal to the amount at the top, but he gained some energy as he traveled due to a higher speed.
Physics
2 answers:
elena55 [62]3 years ago
4 0

Answer:

Just find the formula then you can do it

pickupchik [31]3 years ago
4 0

Answer:

D. The total amount of energy at the bottom is equal to the amount at the top, but he gained some energy as he traveled due to a higher speed.

Explanation:

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A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60
miskamm [114]

Answer:

Answer:196 Joules

Explanation:

Hello

Note:  I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work  is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules

Answer:196 Joules

I hope it helps

5 0
3 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
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