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Maru [420]
3 years ago
12

4. If the Kitty Hawk Flyer has 12 HP, do some research to find out how to express this power in the SI system and report the num

erical value in kilowatts.
Physics
2 answers:
lesya [120]3 years ago
8 0

Product of my research: 1 HP = 746 watts .

12 HP = (12 x 746 W) = 8,952 W

8,952 W = 8.952 kW
Alik [6]3 years ago
4 0

Explanation:

The Kitty Hawk Flyer has 12 HP. We need to express this power in the SI system and in kilowatts.

The SI unit of power is watts and HP is also one of the unit of power. The relationship between the watts and horse power is :

1 HP = 745.7 watts

To convert 12 HP to watts we will use the above relation. So,

12 HP = 8948.4 watts

Also, 1 kilowatts = 1000 watts

So, 8948.4 watts = 8.94 kilowatts

Hence, this is the required solution.

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Answer:

<h2>18 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

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We have the final answer as

<h3>18 N</h3>

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6 0
2 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}

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