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3 years ago

If we decrease the amount of force, and kelp all other factors the same, what will happen to the amount of work?

Physics

1 answer:

Artist 52 [7]3 years ago

8 0

Answer:

Distance will decrease and work will decrease:

F = m a Newton's Second Law

a = F / m decreasing force will decrease acceleration

S = 1/2 a t^2 = 1/2 (F / m) t^2 distance traveled will decrease as force decreases

W = F * S work will decrease as both force and distance decrease

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3 years ago

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A train travels 90 kilometers in 2 hours, and then 66 kilometers in 2 hours. What is its average speed?

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3 years ago

An example of constant velocity

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2 years ago

a bucket of water of mass 20kg is pulled at a constant velocity up to a platform 40 meters above the ground. this takes 10 minut

erma4kov [3.2K]

<span>Taking into account the information above, we know the average mass of the bucket of water may be m=20-5/2=17.5kg. As the bucket of water is pulled at a "constant velocity" the work required to raise the bucket to the platform transformed into the potential energy of the bucket of water. That is why it should be W=mgh=17.5*9.8*40=6860J</span>

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3 years ago

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago