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dmitriy555 [2]
3 years ago
9

A 63.2-kg climber finds herself dangling over the edge of a cliff. Fortunately, she’s connected by a rope of negligible mass to

a 1220-kg rock located 48.6 m from the edge of the cliff. Unfortunately, the ice is frictionless, so the climber accelerates downward. What’s her acceleration, and how much time does she have before the rock goes over the edge?
Physics
1 answer:
wel3 years ago
5 0

Answer:0.535 m/s^2

Explanation:

Given

mass of climber(m_1)=63.2 kg

Distance between rock and cliff=48.6 m

mass of rock(m_2)=1220

let T be the tension in the rope

Thus T-m_1g=m_1a------1

where a is the acceleration of system

Also for Rock

T=m_2a------2

From 1 & 2 we can say that

m_1\left [ g+a\right ]=m_2\left [ a\right ]

g+a=\frac{1220}{63.2}\left [ a\right ]

g+a=19.303 a

g=18.303 a

a=\frac{g}{18.303}

a=0.535 m/s^2

Thus climber is decelerating with 0.535 m/s^2

time to cover 48.6 m

48.6=\frac{0.535t^2}{2}

t=13.47 s

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Answer: 1m/s

Explanation: according to the law of conservation of linear momentum in an isolated system, the momentum of the gun equals that of the bullet.

Mathematically

Mb×Vb = Mg×Vg

Where Mb = mass of bullet = 1/100 = 0.01 kg

Vb = velocity of bullet = 200 m/s

Mg = mass of gun = 2kg

Vg = recoil velocity of gun =?

0.01×200 = 2×Vg

Vg = 0.01×200/2

Vg = 0.01×100

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PLEASE ANSWER QUICK!! A student pushes a wagon full of bricks with a constant force across the ground. Which of
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A

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If you do D, it would slow it down, due to there being more weight but same force.

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jenyasd209 [6]

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3 years ago
The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km.hr. the tires have a diamet
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In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
SOVA2 [1]

Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

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distance traveled in this time

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d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

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1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

a=0.084 m/s^2

5 0
3 years ago
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