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2 years ago

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In a real machine, the work output is always less than the work input. a. True b. False

1 answer:

Artist 52 [7]2 years ago

4 0

<span>In a real machine, the work output is always less than the work input is true. The answer is letter A. it follows the law of entropy where no energy can be converted completely into work. Under this law, Carnot’s theorem states that Carnot’s engine can perform 100% of work. However, no such engine has ever succeeded the conversion of work into 100%. The greatest efficiency so far is at 80%. Because there will always be factors that could affect the conversion of work. </span>

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Crazy boy [7]

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1 year ago

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A 70.0 kg skydiver falls towards the earth. If the force due to air resistance is 0 N, what is the acceleration of the skydiver?

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The gravitational acceleration g = 9.81 m/s²

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3 years ago

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Two light bulbs are installed in two sockets, the sockets are connected in series, and power is then applied to the combination

Arisa [49]

Answer:

False

Explanation:

<em>If one of the bulbs is removed from the series, the other bulb will not come on at all.</em>

This is because the removal of one of the bulbs would interrupt the flow of current though the entire circuit.

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2 years ago

Olivia places her pet frog on a line to observe the frog’s motion. The line is divided into sections that measure 1 centimeter e

aalyn [17]

Answer:

Explanation:

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3 years ago

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A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler

Pepsi [2]

Answer:

$331665750000\ m/s^2$

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = $1.989\times 10^{30}\ kg$

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2$

The gravitational acceleration at the surface of such a star is $331665750000\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s$

The velocity of the object would be 3257806.62409 m/s

8 0

3 years ago