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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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g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor
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R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1,  and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

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R1 = 3.08/0.6

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6 0
3 years ago
A proton initially has v=4.0i^−2.0j^+3.0k^ and then 4.0 s later has v=−2.0i^−2.0j^+5.0k^ (in meters per second). For that 4.0 s,
pishuonlain [190]

Answer:

a_{avg}=-1.5i+0.5k

1.58113883008\ m/s^2

-18.43^{\circ}\ or\ 161.57^{\circ}

Explanation:

u = 4.0i−2.0j+3.0k v = −2.0i−2.0j+5.0k

Average acceleration is given by

a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{-2-4, -2+2, 5-3}{4}\\\Rightarrow a=-1.5i-0j+0.5k

a_{avg}=-1.5i+0.5k

The magnitude is

a_{avg}=\sqrt{(-1.5)^2+0.5^2}\\\Rightarrow a_{avg}=1.58113883008\ m/s^2

The magnitude is 1.58113883008\ m/s^2

The angle is

\theta=tan^{-1}\dfrac{a_z}{a_x}\\\Rightarrow \theta=tan^{-1}\dfrac{0.5}{-1.5}\\\Rightarrow \theta=-18.43^{\circ}\ or\ 161.57^{\circ}

The angle between a_{avg} and the positive direction of the x axis is -18.43^{\circ}\ or\ 161.57^{\circ}

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2 years ago
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