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A small coin of mass m1 is undergoing a uniform circular motion at a velocity v. The radius of the circular path is r. A piece o

GarryVolchara [31]

Answer:

Answere is (D)

Explanation:

Given m1, v and m2

Total circular momentum is conserved: total circular momentum before putty coin collision is equal to total circular momentum after collision.

Let the final circular velocity of the system be V. Initial circular velocity of the putty is zero.

m1v + m2× 0 = (m1 + m2) V

V = m1v/(m1 + m2).

3 0

3 years ago

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

sergejj [24]

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?

5 0

4 years ago

What is the weight in pounds of a 7.0 kilogram bowling ball on earth's surface

hodyreva [135]

Gravitational field strength (GFS) on earth =10 N/kg

Weight = GFS x mass
W = 10 x 7
W = 70N
70N = 15.74 pounds

3 0

3 years ago

Read 2 more answers

Two immersion heaters, A and B, are both connected to a 120.0-V supply. Heater A can raise the temperature of 1.00 L of water fr

Inessa05 [86]

Answer:

Ratio of resistance of heater A to resistance of heater B is 5.80