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mariarad [96]
3 years ago
14

A 8kg cat us running 4 m/s. How much kenetic energy is that

Physics
1 answer:
Flauer [41]3 years ago
3 0
     Using the Definition of Kinetic Energy, we have:

E_{c}= \frac{mv^2}{2}  \\ E_{c}= \frac{8*4^2}{2}  \\ \boxed {E_{c}=64J}
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A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
on june 21 some earth locations have 24 hours of daylight these locations are all between the latitudes of
andrew-mc [135]

66° N and 90° N

the area of the artic circle in the northern hemisphere

4 0
3 years ago
If the speed of sound in air at 20 °C is 342m/s , what will be the increase in speed at 30 °C? with steps
andrew-mc [135]

Answer:

Hmm

Explanation:

I don't know sorry forgive me.

6 0
3 years ago
A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s^2 . She is driving at 20 m/s when su
QveST [7]

Answer:

The car stops after 32.58 m.

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 20 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6 m/s²

Time taken by the car to stop

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20}{-6}\\\Rightarrow t=3.33\ s

Total Time taken by the car to stop is 0.5+3.33 = 3.83 s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=20\times 3.83+\frac{1}{2}\times -6\times 3.83^2\\\Rightarrow s=32.58\ m

The car stops after 32.58 m.

Distance between car and obstacle is 50-32.58 = 17.42 m

4 0
3 years ago
The four major Earth systems are the atmosphere, lithosphere, hydrosphere, and geosphere. true and false
love history [14]
False its atmosphere, lithosphere, hydrosphere, and boisphere
5 0
3 years ago
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