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Marizza181 [45]
2 years ago
8

A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2 m/s to 4 m/s?

Physics
2 answers:
valentinak56 [21]2 years ago
5 0
Momentum = (mass) x (speed)

Choice-B is the correct one.
mash [69]2 years ago
3 0

Answer:

B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s

Explanation:

The momentum of an object is given by the product between its mass (m) and its velocity (v):

p=mv

Let's apply this formula to calculate the initial momentum and final momentum of the ball:

- initial momentum:

p_i = m v_i = (6 kg)(2 m/s)=12 kg m/s

- Final momentum:

p_f = m v_f = (6 kg)(4 m/s)=24 kg m/s

So, the correct answer is

B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s

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Answer:

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Explanation:

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Explanation:

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2 years ago
A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95
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Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as  y=l'-l=41-30=11m

the mass is m=95 kg

the  force is  F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of  k is given by equating the energy at the equilibrium point which is given as

U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m

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x=\dfrac{F}{k}\\

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3 years ago
Question 15 plz picture above
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