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puteri [66]
3 years ago
8

What is the kinetic energy of a 5-kg object moving at 7 m/s?

Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0
To answer this question, we would need the formula for the kinetic energy which is Kinetic Energy = ½ x m x v^2

Where the following means: m is the mass of the object and v is the velocity

So At 7 m/s: Kinetic Energy = ½ x 5 x 7^2 = 122.5 J is the answer
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Which reading strategy helps you to read and understand a passage quickly by making use of your previous knowledge?
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Taking notes,because you can read back on what you learned
4 0
3 years ago
Water (density = 1 ´ 103 kg/m3) flows at 10 m/s through a pipe with radius 0.030 m. the pipe goes up to the second floor of the
Hatshy [7]

density of water = 1000 kg/m^3

velocity of flow = 10 m/s

radius of pipe = 0.030 m

Height of second floor = 2 m

Now we can use here Bernuoli's Equation to find the speed of water flow at second floor

P_1 + 1/2\rho v_1^2 + \rho g h_1= P_2 + 1/2 \rho v_2^2 + \rho g h_2

P + 1/2 * 1000 * 10^2 + 1000* 9.8 * 0 = P + 1/2 * 1000 * v^2 + 1000*9.8*2

v = 7.8 m/s

Now in order to find the radius of pipe we can use equation of continuity

A_1 v_1 = A_2 v_2

\pi *0.030^2 * 10 = \pi * r^2 * 7.8

r = 0.034 m

So radius of pipe at second floor is 0.034 meter

3 0
3 years ago
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
If an objects speed is constant and the direction is in a straight line what type of velocity does the object have
storchak [24]
That's unaccelerated motion, and constant velocity.
3 0
3 years ago
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A sprinter who is running a 200-m race travels the second 100 m in much less time than the first 100m because
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Answer:

He is warmed up now

Explanation:

His muscles are better and stretched now

3 0
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