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Tresset [83]
2 years ago
7

a 50kg person is standing still in ice skates throws their 2.0kg helmet to the right at 25m/s while on a friction less surface.

what is the final speed of the person
Physics
1 answer:
Roman55 [17]2 years ago
8 0

Answer:

Vp = 1 [m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of the amount of movement. In the same way, analyze the before and after of the actions.

<u>The moment before</u>

The 50kg person is still hold (no movement) with the 2kg helmet

<u>The moment after</u>

The helmet moves at 25[m/s] in one direction, the person moves in the opposite direction, due to the launch of the helmet.

In this way we can apply the principle of conservation of movement, expressing the before and after. To the left we have the before and to the right of the equal sign we have the after.

Σm*V1 = Σm*V

where:

m = total mass = (2 + 50) = 52[kg]

V1 = velocity before the lunch = 0

(50 + 2)*V1 = (25*2) - (Vp*50)

0 = 50 - 50*Vp

50 = 50*Vp

Vp = 1 [m/s]

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(a) The y-component or vertical velocity is calculated using:
Vy = Vsin(∅)

(b) The x-component or horizontal velocity is calculated using:
Vx = Vcos(∅)
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2 years ago
The image on the left shows the charges on a balloon after it’s been rubbed with a wool cloth. The image on the right is a piece
taurus [48]

When wool is rubbed with a balloon, the wool is left with a positive charge as electrons have travelled from the wool to the balloon which means the balloon now has a negative charge.

Now that the balloon has a negative charge, you need to know:
The tissue paper originally contains electrons and protons
The fact that the balloon has a negative charge, it will ATTRACT protons because protons are POSITIVE and electrons are NEGATIVE.
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3 0
2 years ago
Read 2 more answers
Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what
hoa [83]

Answer:

<h2>187,500N/m</h2>

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}

The value of the spring constant is 187,500N/m

7 0
2 years ago
An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a h
marin [14]

Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

K=\frac{1}{2}mv^{2}

For the object thrown in the air:

K=\frac{1}{2}.2.[v(t)]^{2}

K=(-9.8t+24)^{2}

K=96.04t^{2}-470.4t+576

Kinetic energy of the object as a function of time: K=96.04t^{2}-470.4t+576

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

U=mgh

For the object thrown in the air:

U=9.8.2.h(t)

U=9.8.2.(-4.9t^{2}+24t+60)

U=-96.04t^{2}+470.4t+1176

Potential energy as function of time: U=-96.04t^{2}+470.4t+1176

Total kinetic and potential energy, also known as mechanical energy is

TME = 96.04t^{2}-470.4t+576 + (-96.04t^{2}+470.4t+1176)

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.

6 0
2 years ago
The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at
elena-14-01-66 [18.8K]

The position function x(t) of a particle moving along an x axis is x=4.00 - 6.00t^2

a) The point at which particle stop, it's velocity = 0 m/s

  So dx/dt = 0

        0 = 0- 12t = -12t

  So when time t= 0, velocity = 0 m/s

    So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

 x=4.00 - 6.00*0^2 = 4m

SO at x = 4m the particle is (momentarily) stop

c) We have x=4.00 - 6.00t^2

   At origin x = 0

  Substituting

         0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}

         t = 0.816 seconds or t = - 0.816 seconds

So when  t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

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