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3 years ago

6

Assume an 8-kg bowling ball moving at 2m/s bounces off a spring at the same speed that it had before bouncing.What is the moment

um of the recoil?

1 answer:

aliya0001 [1]3 years ago

5 0

Answer:-16kgm/s

Explanation:

-mv=-2×8=-16kgm/s

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Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

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$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

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nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

F-400 = 1.5 x 10³x18 =27000 ,

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Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

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