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brilliants [131]
3 years ago
6

Please help??!

Physics
2 answers:
lina2011 [118]3 years ago
8 0

Answer:

Windy, whooping weather

Explanation:

thats the answer

KatRina [158]3 years ago
4 0

Answer: 4. Windy, whooping weather

Explanation: b/c it is!

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What happens to the temperature of a gas when it is compressed? a the temperature does not change. b the temperature increases.
nasty-shy [4]
I think the temperature increases
5 0
4 years ago
Read 2 more answers
A car travels a distance of 320 km in 4 hours. What is your average speed in meters per second?
Andreas93 [3]

Answer:

22.2 m/s

Explanation:

First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.

Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.

The average speed can be found by using the equation \frac{distance}{time}. After substitution, this gives the fraction \frac{320 000}{14 400}, which reduces to 22 \frac{2}{9} m/s, or about 22.2 m/s.

4 0
4 years ago
1. A cyclist accelerates from 0 m/s to 9 m/s in 3 seconds. What is his<br>acceleration?​
Lapatulllka [165]

The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

Answer:

Explanation:

Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.

Then acceleration = change in velocity/Time.

Acceleration = \frac{Change in velocity}{Time taken}

Acceleration = (9-0)/3=9/3=3 m/s².

So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

3 0
3 years ago
Determine whether the center of mass of the system consisting of the earth and moon lies inside or outside the earth. Assume tha
tester [92]

Answer:

R_cm = 4.66 10⁶ m

Explanation:

The important concept of mass center defined by

         R_cm = 1 / M   ∑  x_i m_i

where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

       R_cm = 1 / M (m_earth 0 + m_moon d)

the total mass is

      M = m_earth + m_moon

     

the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

      M = 5.98 10²⁴ + 7.35 10²²

      M = 6.0535 10₂⁴24 kg

we substitute

      R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

      R_cm = 4.66 10⁶ m

4 0
3 years ago
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the
Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0
4 years ago
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