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zloy xaker [14]
3 years ago
8

A robotic arm lifts a stack of cafeteria trays weighing 62 newtons (N). It moves the trays 2 meters (m). How much work has the r

obotic arm performed?
Physics
1 answer:
tigry1 [53]3 years ago
7 0

work done = force * distance

work = 62 * 2 = 124 j

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How much displacement will a spring with a constant of 120N / m achieve if it is stretched by a force of 60N?
Andrew [12]

Answer:

Explanation:

There's a formula for this:

F = k*displacement

F being force, k being the spring constant, and displacement being the change in x

We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters

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In operant conditioning, generalization occurs when
Vitek1552 [10]
Typically occurs when we associate things to other things that look alike. We see that in many experiments, specifically “Little Albert” who was conditioned to be afraid of rats but later was afraid of anything that resembled that of a rat.
Hope this helps!
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3 years ago
The forces on a car are balanced.state and explain the resultant force on the car
Vanyuwa [196]

there will no resultant force

Explanation:

this is because if the forces are balanced on opposite direction. then they cancel each other out

5 newton's ---------> <--------- 5 newton's

then both forces will cancel each other out as a result there is no resultant force and the newton's laws states that if there is no resultant the object will continue in its state of rest (remains there) or it will in continue in its uniform motion in a straight line.

I hope you understand,

5 0
3 years ago
A uniformly charged, straight filament 5.10 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder
Ratling [72]

Answer:

70509.8039216 N/C

Explanation:

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q = Charge = 2.00 µC

l = Length of filament = 5.1 m

r = Radius of cylinder = 10 cm

\lambda=\dfrac{q}{l}

Electric field is given by

E=\dfrac{2k\lambda}{r}\\\Rightarrow E=\dfrac{2\times 8.99\times 10^9\times \dfrac{2\times 10^{-6}}{5.1}}{10\times 10^{-2}}\\\Rightarrow E=70509.8039216\ N/C

The electric field at the surface of the cylinder is 70509.8039216 N/C

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