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3 years ago

How is an object’s position in the solar system affect its motion and temperature?

Physics

1 answer:

SCORPION-xisa [38]3 years ago

8 0

Answer:

it will be base on the climate change.

Explanation: so let just say tomorrow it will be cold so the atmosphere will circle on the coolest atmosphere away from the sun.

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ILL GIVE BRAINLIEST!! please help and answer correctly!

Hunter-Best [27]

Answer: B

Action and reaction forces act on the same object and go to the same direction

6 0

3 years ago

What does the diagram show about phases and the phase of the substance as it is heated? Check all that apply.

Romashka [77]

Answer:

2 and 4

Explanation:

right on edge

7 0

3 years ago

Read 2 more answers

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00×105 Pa

Alborosie

Explanation:

In the first case , the volume is kept constant and pressure varies .

In isothermal process , the work done

W₁ = V x ΔP

here V is the volume of gas and ΔP is the change in pressure

Thus W₁ = 0

Because there is no change in volume , therefore displacement is zero .

In second case pressure is constant , but volume changes

Thus W₂ = P x ΔV

here P is the pressure and ΔV is the change in volume

Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J

The total work done W = - 2 x 10⁴ J

Because the work done in compression is negative .

7 0

4 years ago

A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00 m above the ground. At th

Elis [28]

Answer:

<u>The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.</u>

Explanation:

The motions of the two water ballons are ruled by the kinematic equations:

$y=y_0+V_0t-gt^2/2$

We are only interested in the vertical motion, so that equation is all what you need.

<u>1. Water ballon is thrown horizontally at sped 2.00 m/s.</u>

The time the ballon takes to hit the ground is independent of the horizontal speed.

Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.

Then:

$y=y_0+V_0t-gt^2/2\\ \\ 0=6.00m-9.8\frac{m}{s^2} t^2/2\\ \\ t=\sqrt{2\times6.00m/9.8\frac{m}{s^2}}\\\\ t=1.11s$

<u>2. Water ballon thrown straight down at 2.00 m/s</u>

Now the initial vertical speed is 2.00 m/s down. So, the equation is:

$0=6.00m-2.00\frac{m}{s}t-9.8\frac{m}{s^2}t^2/2\\ \\ 4.9t^2+2t-6=0\\ \\ t=0.92s$

To solve the equation you can use the quadratic formula.

$t=\frac{-2+/-\sqrt{2^2-4(4.9)(-6)} }{2(4.9)}\\ \\ t=-1.33\\ \\ t=0.92$

You get two times. One of the times is negative, thus it does not have physical meaning.

<u>3. Conclusion:</u>

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

5 0

3 years ago

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length t

gladu [14]

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball : 16.3 g

Predicted range : 0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

$t=\sqrt{\frac{2H}{g}}$

$v_1t= 0.35$ ...........(i)

$v_0t= 1.09$ ...........(ii)

Dividing the equation (ii) by (i)

$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$

∴ $v_0=3.11 \ v_1$

Now loss of energy = change in the kinetic energy

$W=\frac{1}{2} m [v_0^2-v_1^2]$

$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$

$W=7.307\times 10^{-3} \ v_0^2$

If f is average friction force, then

(f)(L) = W

(f) (1) = $7.307\times 10^{-3} \ v_0^2$

(f) = $7.307\times 10^{-3} \ v_0^2$

3 0

3 years ago