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andrew-mc [135]

3 years ago

8

A force of 6.0 Newtons is applied to a block at rest on a horizontal frictionless surface over a 7.0 meter span. How much energy

is gained by the block?
a) 3.0 m/s

b) 7.0 m

c) 42 J

d) 6.0 N

1 answer:

lys-0071 [83]3 years ago

8 0

Answer:

C. 42J

Explanation:

Given data

Force= 6N

Distance= 7m

Required

The energy gained

Work= Force* Distance

Work= 6*7

Work= 42Joules

Hence the option c 42J is correct

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3 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface

Keith_Richards [23]

Answer:

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Explanation:

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Acceleration due to gravity is given by

$g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}$

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$v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s$

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3 0

3 years ago

Read 2 more answers

A 2800 kg truck moving at 12 m/s to the right hits a stopped 1100 kg car. What is the combined velocity the moment they stick to

leva [86]

Answer:

The combined velocity is 8.61 m/s.

Explanation:

Given that,

The mass of a truck, m = 2800 kg

Initial speed of truck, u = 12 m/s

The mass of a car, m' = 1100 kg

Initial speed of the car, u' = 0

We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.

$m_1v_1+m_2v_2=(m_1+m_2)V\\\\V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\V=\dfrac{2800\times 12+0}{2800+1100}\\\\V=8.61\ m/s$

So, the combined velocity is 8.61 m/s.

5 0

3 years ago