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3 years ago

IF EARTH HAD NO ATMOSPHERE, WOULD A FALLING OBJECT EVER REACH TERMIONAL VELOCITY?

1 answer:

3 0

No. What most people call 'terminal velocity' is the speed of the falling
object when the downward force of gravity is equal to the upward force
of air resistance. At that speed, the vertical forces on the object are
balanced, so it stops accelerating, and falls at a constant speed.

If there were no atmosphere, there would be no upward force due to
air resistance. The falling object would continue to accelerate all the
way down until it went 'splat'.

This is exactly the situation for meteoroids or asteroids falling onto the Moon.

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A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains

IrinaK [193]

Answer:

503°C

Explanation:

According to the given situation, the computation of the final temperature is shown below:

In this question we use the law of ideal gas i.e

pV = nRT

i.e

$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$

Therefore

$T_2 = T_1 (\frac{p_2}{p_1}) (\frac{V_2}{V_1})$

$= 300\ k (\frac{2.72 \times 10^{6} Pa + 1.01 \times 10^{5} Pa}{1.01 \times 10^{5} Pa})(\frac{46.2 cm^3}{499 cm^3})$

= 776 k

= (776 - 273)° C

= 503°C

Therefore the final temperature is 503°C

We simply applied the above formulas so that the final temperature could arrive

5 0

3 years ago

A water rocket has a mass of 0.8kg and is launched in a school playground with an inital upwards force of 12newtons. what is the

Darina [25.2K]

Weight = (mass) x (gravity).
It always acts downward.

On Earth, the acceleration of gravity is 9.807 m/s².
On the Moon, the acceleration of gravity is 1.623 m/s².

On Earth, the rocket's weight is (0.8kg) x (9.8 m/s²) = 7.84 newtons

On the Moon, the rocket's weight is (0.8kg) x (1.62 m/s²) = 1.3 newtons

The force of the rocket engine acts upward.
Its magnitude is 12 newtons. (From the burning chemicals.
Doesn't depend on local gravity. Same force everywhere.)

Now we have all the data we need to mash together and calculate the
answers to the question. You might choose a different method, but the
machine that I have selected to do the mashing with is Newton's 2nd law
of motion:

   Net Force = (mass) x (acceleration).

Since the question is asking for acceleration, let's first solve Newton's law
for it. Divide each side by (mass) and we have

   Acceleration = (net force) / (mass) .

On Earth, the forces on the rocket are

      (weight of 7.84 N down) + (blast of 12 N up) = 4.16 newtons UP (net)

   Acceleration = (4.16 newtons UP) / (0.8 kg) = 5.2 m/s² UP .

On the moon, the forces on the rocket are

   (weight of 1.3 N down) + (blast of 12 N up) = 10.7 newtons UP (net)

   Acceleration = (10.7 newtons UP) / (0.8 kg) = 13.375 m/s² UP

7 0

3 years ago

Sarah drives her car with a constant speed of 48 miles per hour. How far can she travel in 60 minutes?

Setler79 [48]

She can dive 58 miles per hour

4 0

3 years ago

Read 2 more answers

I need help ASAP I need to get this right plz plz plz!!!!!

Rina8888 [55]

The atomic nucleus consists of protons and neutrons.

Additional information:

Protons are positively charged particle and neutrons are negatively charged particle.

Hope this will help u...:)

3 0

3 years ago

Read 2 more answers

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

3 years ago