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Doss [256]
3 years ago
9

A father racing his son has 1/4 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.2 m

/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?
Physics
1 answer:
sergejj [24]3 years ago
5 0

Answer:

Explanation:

KE_s: Kinetic Energy Son

KE_f: Kinetic Energy Father.

Relationship

KE_f: =  (1/4) KE_s

m_s: = (1/3) m_f

v_f: = velocity of father

v_s: = velocity of the son

Relationship

1/2 mf (v_f + 1.2)^2 = 1/2 m_s (v_s)^2      Multiply both sides by 2.

mf (v_f + 1.2)^2 = m_s * (v_s)^2               Substitute for the mass of the m_s

mf (v_f + 1.2)^2 = (m_f/3) * (v_s)^2         Divide both sides by father's mass

(v_f + 1.2)^2 = 1/3 * (v_s)^2                      multiply both sides by 3

3*(v_f + 1.2)^2 = (v_s) ^2                         Take the square root both sides

√3 * (v_f + 1.2) = v_s

Note

  • You should work your way through all the cancellations to find the last equation shown about
  • We have another step to go. We have to use the first relationship to get the final answer.

KE_f = (1/4) KE_s                                                  Multiply by 4

4* KE_f = KE_s                                                     Substitute (again)

4*(1/2) m_f (v_f + 1.2)^2 = 1/2* (1/3)m_f *v_s^2   Divide by m_f

2* (v_f + 1.2)^2 = 1/6 * (v_s)^2                              multiply by 6

12*(vf + 1.2)^2 = (v_s)^2                                        Take the square root

2*√(3* (v_f + 1.2)^2) = √(v_s^2)

2*√3 * (vf + 1.2) = v_s

Use the second relationship to substitute for v_s so you can solve for v_f

2*√3 * ( v_f + 1.2) = √3 * (v_f + 1.2)                     Divide by sqrt(3)

2(v_f + 1.2) = vf + 1.2

Edit

2vf + 2.4 = vf + 1.2

2vf - vf + 2.4 = 1.2

vf = 1.2 - 2.4

vf = - 1.2

This answer is not possible, but 2 of us are getting the same answer. The other person is someone whose math I would never question. She rarely makes an error. And I do mean rarely. Could you check to see that you have copied this correctly?

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leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

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\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

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3 years ago
Which BEST describes the physical properties of the Earth’s core?
Sonja [21]

C) a solid lower part and a liquid upper part

Explanation:

The physical nature of the earth's core is made up of a solid lower part and a liquid upper part.

The core is the innermost part of the earth and it is made up of metallic minerals.

  • It has the highest temperature and pressure of all the layers of the earth.
  • The core is divided into two. Outer and inner core.
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learn more:

Crust brainly.com/question/10537829

#learnwithBrainly

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A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings m
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Answer:

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Explanation:

m_1 = Mass of bullet = 19 g

m_2 = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

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u = Velocity of bullet

Combined velocity of bullet and bob is given by

v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s

As the momentum is conserved

m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s

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A 3.0 kg box is suspended by a series of ropes as shown below. The tension force in the horizontal rope is -40 N. What is the te
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Answer:

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The horizontal component of the forces:

F₁ + F₂ = -40N + F₂ = 0

F₂ = 40N

The vertical component of the forces:

F₁ + F₂ - mg = 0 + F₂ - mg = 0

F₂ = mg

If I assume the gravitational constant g = 10 m/s²:

F₂ = (3 kg) * (10 m/s²) = 30N

Adding the horizontal and vertical components of the force F₂:

F₂ = √((40N)² + (30N)²) = 50N

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the answer is

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