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3 years ago

What is it called when a skydiver is no longer accelerating downward?

Physics

1 answer:

raketka [301]3 years ago

4 0

Explanation:

Once the force of air resistance is as large as the force of gravity, a balance of forces is attained and the skydiver no longer accelerates. The skydiver is said to have reached a terminal velocity.

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Which of the following contains the most thermal energy?

vodka [1.7K]

The answer would be c.

6 0

4 years ago

A plastic light pipe has an index of refraction of 1.48. For total internal reflection, what is the minimum angle of incidence i

Rama09 [41]

Answer:

a) 42.52°

b) 63.98°

Explanation:

Refractive index of pipe = 1.48 = n₂

Refractive index of air = 1.0003 = n₁

$\theta_r=sin^{-1}\frac{n_1}{n_2}\\\Rightarrow \theta_r=sin^{-1}\frac{1.0003}{1.48}\\\Rightarrow \theta_r=sin^{-1}0.676\\\Rightarrow \theta_r=42.52^{\circ}$

∴ Minimum angle of incidence is 42.52°

Refractive index of water = 1.33 = n₁

$\theta_r=sin^{-1}\frac{n_1}{n_2}\\\Rightarrow \theta_r=sin^{-1}\frac{1.33}{1.48}\\\Rightarrow \theta_r=sin^{-1}0.899\\\Rightarrow \theta_r=63.98^{\circ}$

∴ Minimum angle of incidence is 63.98°

5 0

4 years ago

The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s.

Aleks [24]

By definition, centripetal acceleration is given by:

$a = \frac{v ^ 2}{r}$

Where,

v: tangential disk speed

r: disk radius

Substituting values in the given equation we have:

$a =\frac{3.3^2}{0.13}\\a = 83.76923077$

Rounding the result we have:

$a = 83.8 \frac{m}{s^2}$

Answer:

The centripetal acceleration of the disc edge in m/s^2 is:

$a = 83.8 \frac{m}{s^2}$

3 0

4 years ago

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Which terms describe space objects that are small chunks of rock and debris that are smaller than 1 km?

UkoKoshka [18]

I would say that it is Meteors, Meteorites, and comets. (METEORS may be wrong)

5 0

3 years ago

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Heat is allowed to flow from the heat source of a heat engine at 425 K to a cold sink at 313 K. What is the efficiency of the he

olga2289 [7]

I assume here that the engine operates following a Carnot cycle, which achieves the maximum possible efficiency.

Under this assumption, the efficiency of the engine (so, the efficiency of the Carnot cycle) is given by
$\eta = 1- \frac{T_{cold}}{T_{hot}}$
where
$T_{cold}$ is the cold temperature
$T_{hot}$ is the hot temperature

For the engine in our problem, the cold temperature is 313 K while the hot temperature is 425 K, so the effiency of the engine is
$\eta=1- \frac{313 K}{425 K}=0.264 = 26.4 \%$

3 0

4 years ago

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