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3 years ago

What is the work done by 20 Newton force applied at an angle of 45.0° to move a box a horizontal distance of 40 meters

Physics

2 answers:

ludmilkaskok [199]3 years ago

4 0

Hope this helps! let me know if you need clarification

zzz [600]3 years ago

3 0

Answer:

5.6 × 10^2 joules

Explanation:

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Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 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>

5 0

3 years ago

Which of the following is not an example I work?? Some please help

n200080 [17]

B. picking up a box off the floor

8 0

3 years ago

Read 2 more answers

What distance separates Earth and the Sun? A 300,000 km per second B Eight light-minutes C 149 million light-seconds D One light

lina2011 [118]

Answer:

B Eight light-minutes

Explanation:

In the case when the distance separated earth and the sun so here we orbit the sun for a 150 million km distance and the light moves would be 300,000 kilometers per second

Now divide this

= 150 million ÷ 300,000 kilometers per second

= 500 seconds

This 500 seconds represent 8 minutes and 20 seconds

Hence, option B is correct

5 0

3 years ago

Read 2 more answers

If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.130 m from equilibrium?

puteri [66]

Answer:

The speed is 1.52 m.