All categories

3 years ago

True or false the earth loses energy to space only at night

1 answer:

6 0

Well first of all, there's nothing that "the Earth" does at night,
because 'the Earth' is never completely 'night'. "The Earth"
is always half night and half day.

Now, your question probably means: The Earth loses energy
to space only from the part that's dark, where the sun don't shine.

That's false. The Earth is warm and space is cold, so there's
heat radiating into space from every place on Earth all the time.
Fortunately for us, the sun shines on every place on Earth for
50% of the time, and pours down as much energy as radiates
away into space.

And more. That, and the fact that we've been pumping stuff into
the air for the past 300 years that makes it harder for heat to get
away into space, are the reasons why the Earth is getting hotter.
Can it keep going ? Sure. But past a certain point, we can't
live in it, so we start to decrease, and we eventually go extinct.

You might be interested in

Calculate the pressure on the bottom of a swimming pool 3. 5 m deep. How does the pressure compare with atmospheric pressure, 10

Nostrana [21]

Answer:

1.343 atm

Explanation:

The mass of water above 1 square meter of swimming pool bottom is ...

M = (3.5 m)·(1000 kg/m^3) = 3500 kg/m^2

Then the force exerted by the water on the pool bottom is ...

F = Mg = (3500 kg/m^2)(9.8 m/s^2) = 34300 N/m^2 = 34300 Pa

Compared with atmospheric pressure, this is ...

34,300/10^5 = 0.343 . . . . atmospheres

Added to the atmospheric pressure on the water's surface, the total pressure on the pool bottom is 1.343 atmospheres.

5 0

4 years ago

A concert loudspeaker suspended high off the ground emits 33.0 W of sound power. A small microphone with a 0.600 cm2 area is 52.

Novay_Z [31]

Answer:

The sound intensity at the position of the microphone is $9.71\times10^{-4} W/m^{2}$

Explanation

Sound intensity is given by the formula

$I=\frac{P}{A}$

Where $I$ is the sound intensity, $P$ is the power and $A$ is the area.

Since the loudspeaker radiates sound in all directions, we have a spherical sound wave where the radius r is the distance of the microphone from the speaker.

∴ $A$ is given by $4\pi r^{2}$ where $r$ is the radius

From the question, $P$ = 33.0W, $r$ = 52.0m

$I=\frac{P}{A} = \frac{P}{4\pi r^{2} }$

$I = \frac{33.0}{4\pi \times (52.0)^{2} }$

∴ $I = 9.71\times10^{-4} W/m^{2}$

Hence, the sound intensity at the position of the microphone is 9.71 × 10⁻⁴ W/m²

7 0

3 years ago

a 200 kg crate is pushed horizontally with a force of 700 N. If the coefficient of friction is 0.2 calculate the acceleration of

Crazy boy [7]

Answer:

$a=1.54\ m/s^2$

Explanation:

<u>Net Force</u>

The Second Newton's law states that an object acquires acceleration when an external unbalanced net force is applied to it.

That acceleration is proportional to the net force and inversely proportional to the mass of the object.

It can be expressed with the formula:

$\displaystyle a=\frac{F_n}{m}$

Where

Fn = Net force

m = mass

The m=200 kg crate is pushed horizontally with a force Fa=700 N. The friction force opposes motion and a horizontal net force appears causing the acceleration.

The forces on the vertical direction are in balance since the crate does not accelerate in that direction, thus the weight and the normal force are equal:

N = W = mg

The friction force can be calculated by using the coefficient of friction μ:

$F_r=\mu N$