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soldi70 [24.7K]
3 years ago
13

A 20 N force causes an impulse of 500 N*s. How long was the force applied for?

Physics
1 answer:
yKpoI14uk [10]3 years ago
3 0

Answer:

25sec

Explanation:

Impulse = Force * time (period)

therefore,

500N*s= 20N * x(secs)

divide both sides by 20

results in  500/20=x

x= 25 secs

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A human hair is approximately 56 µm in diameter.
Ann [662]

Answer:

The diameter is 0.000056 m

Explanation:

Lets explain the relation between the meter and the micrometer

1 Meter is equal to 1000000 (one million) micrometers

1 micrometer = \frac{1}{1000000}=\frac{1}{10^{6}}=10^{-6}

The symbol of the meter is m

The symbol of micrometer is μm

A human hair is approximately 56 µm in diameter

We need to express this diameter in meter

To do that we divide this number by 1,000,000 or multiply it by 10^{-6}

→ \frac{56}{1000000}=0.000056  56 µm = 0.000056 m

→ OR

→ 56*10^{-6}=0.000056

→ 56 µm = 0.000056 m

<em>The diameter is 0.000056 m</em>

4 0
4 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
3 years ago
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