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Marianna [84]
3 years ago
9

The image shows the path of a ball from the time it’s thrown to the time it lands on the ground. Determine the kind of energy th

e ball has at each position. (PE stands for gravitational potential energy, and KE stands for kinetic energy.)
PE
neither
PE and KE

Physics
1 answer:
LenaWriter [7]3 years ago
4 0

At Position 1: The ball has both PE and KE. (The ball is at a height from ground and velocity is given to the ball)

At Position 2: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.

At Position 3: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 4: The ball has only PE. (The ball is at a height from ground and velocity is zero.)

At position 5: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 6: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 7: The ball has only KE (Just before hitting the ground). After landing the ball have neither of the energy.

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Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, N_P = 50 turns

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the secondary load resistance, R_S = 250 Ω

Determine the turns ratio;

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Now, determine the reflected resistance in the primary winding;

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Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

coefficient of friction: \mu =0.27

and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

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so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

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now let's analyze the horizontal forces

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and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

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