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3 years ago

12

Please help!!

1 answer:

Elina [12.6K]3 years ago

8 0

Answer:

C. 4.0 J

Explanation:

Work is equal to the change in kinetic energy or kinetic energy final - kinetic energy initial. 1/2 mv^2 is the equation for kinetic energy. SInce the mass is the same, you don't have to include it when solving. 1/2 times 1^2 = 1/2. 1/2 times 3^2 = 4.5. 4.5 - 0.5 = 4 J

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3 years ago

Please help!! giving a lot of points

Readme [11.4K]

Question 1.

mass = 4500 kg
potential energy (p.e) = 67500 J

now, we know :

=》

$p.e = mgh$

=》

$67500 = 4500 \times 10 \times h$

=》

$67500 = 45000 \times h$

=》

$h = \dfrac{67500}{45000}$

=》

$h = 1.5 \: m$

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Question 2.

mass = 4500 kg
kinetic energy = 63000 j

we know,

=》

$k.e = \dfrac{1}{2} mv {}^{2}$

=》

$63000 = \dfrac{1}{2} \times 4500 \times {v}^{2}$

=》

${v}^{2} = \dfrac{63000 \times 2}{4500}$

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${v}^{2} = 28$

=》

$v = \sqrt{28}$

=》

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or

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$5.29 \: \: ms {}^{ - 1}$

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3 years ago

Calculate the missing angles in the following diagram if the index of refraction is 1.00

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The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const

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NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

$Power = \frac{Energy}{time}$

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, $s = ut + 0.5at^{2}$

The drill head starts from rest, u = 0 m/s

$5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s$

Power, P = E/t

P = 3920/0.0.943

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But Efficiency, E = 0.65

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6 0

3 years ago

An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici

Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

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From first law of thermodynamics

Q₁ = Q₂ +W

W=Work out put

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W= 1210 J

The efficiency given as

$\eta=\dfrac{W}{Q_1}$

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η = 69.18 %

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$P=\dfrac{W}{t}$

$P=\dfrac{1210}{0.305}\ W$

P=3967.21 W

8 0

3 years ago