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scZoUnD [109]
3 years ago
9

PLZ HELP IM REALLY CONFUSED a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, the

ball (2kg) is traveling 10m/s. Whats the ball's kinetic energy just as it leaves the boy's hand? What is the potential energy of the ball when it reaches the highest point
Physics
1 answer:
erastovalidia [21]3 years ago
3 0

Answer:

K.E = 100 J

Final P.E = 100 J

Explanation:

The kinetic energy of any object can be given by the following formula:

K.E = (\frac{1}{2})mv^{2}

where,

K.E = Kinetic Energy

m = mass of ball = 2 kg

v = speed of ball

Initially, v = 10 m/s. Therefore, the initial K.E is given as:

K.E = (\frac{1}{2})(2\ kg)(10\ m/s)^{2}

<u>K.E = 100 J</u>

Now, at the highest point the K.E of the ball becomes zero. because the ball stops for a moment at the highest point and its velocity becomes zero. So, from Law of Conservation of energy:

Initial K.E + Initial P.E = Final K.E + Final P.E

Initial P.E is also zero due to zero height initially.

K.E + 0 = 0 + Final P.E

<u>Final P.E = 100 J</u>

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a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature
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Answer:

Explanation:

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mars1129 [50]

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A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
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(a) -83.6 Hz

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Here we have

f = 350 Hz

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v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

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(b) 0.865 m

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Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

5 0
3 years ago
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