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Serggg [28]
3 years ago
14

A person applies a force of 67 N for a distance of 11 m to push a desk across the floor. How much work is done

Physics
1 answer:
lbvjy [14]3 years ago
7 0

L=F*d=67*11=737 Joule

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Heisenberg's Uncertainty Principle states that we cannot simultaneously measure both the position and momentum of an object bett
ankoles [38]

For this specific problem, the photons have been localized to D<span>x = </span>0.027m uncertainty. I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.

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3 years ago
A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves
expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

q_1=3.4\mu C=3.4\times 10^{-6} C

1\mu C=10^{-6} C

q_2=-4.9\mu C=-4.9\times 10^{-6} C

x_1=0.125,y_1=0

x_2=0.280,y_2=0.235

We have to find the work done by the electric force on the moving point charge.

r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125

r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366

Work done,W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})

Where k=9\times 10^9

Using the formula

W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})

W=-0.79 J

5 0
4 years ago
What's is meant by relative motion?​
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4 0
3 years ago
Read 2 more answers
can someone describe how someone is moving-- have examples of positive velocity negative velocity and acceleration and the perso
Rama09 [41]

Assume the motion when you are in the car or in the school bus to go to the school.


To describe the motion the first thing you need is a point of reference. Assume this is your house.


This should be a description:

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  • The car starts moving from rest, gaining speed, accelerating. You start to move away from your house, with a positive velocity (from you house to your school) and positive acceleration (velocity increases).
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  • When the car approaches the school, the driver starts to slow down. Then, you speed is lower but yet the velocity is positive, as you are going in the same direction. The acceleration is negative because it is in the opposite direction of the motion.
  • When the car stops, you are again at rest: zero velocity and zero acceleration.
  • In all the path your velocity was positive, constant at times (zero acceleration) and variable at others (accelerating or decelerating).
  • When you comeback home, then you can start to compute negative velocities, as you will be decreasing the distance from your point of reference (your house).


4 0
3 years ago
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