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Lesechka [4]
4 years ago
9

A car takes 8s to increase its velocity from 10m/s to 30m/s.what is its average acceleration ?

Physics
1 answer:
DENIUS [597]4 years ago
7 0
Acceleration is 2.5 m/s^2V1-V0/t= a so 30 m/s - 10 m/s / 8s = 2.5 m/s^2
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It's the law.  Don't be bashful.
5 0
3 years ago
Need help with science pls help me!! 40 points
sp2606 [1]

Answer: a. Regularly, you rub your hands for warmth.

b. Some examples are friction from electricity going through a wire or friction from rubbing your hand together.

c. Without friction, we would fall every time we walked. Because there is friction, we can walk normally. Another example is a ball moving across the field. Without friction, the ball would not slow down.

Explanation:

Overall, friction is important to us because we depend on it.

3 0
2 years ago
Read 2 more answers
How are sunspots related to prominences and solar flares?
solniwko [45]
<h2><u> Answer:</u></h2><h2><u /></h2>

The Sun is a star that experiences changes in its activity in a cycle of about 11 years, known as the solar cycle. During that time, the amount of sunspots on its surface may vary.

But first it is necessary to define what a sunspot is.

A <u>sunspot </u>is a region of the Sun that has a lower temperature than its surroundings, and with an intense magnetic activity. It consists of a dark central region, called <em>"umbra"</em>, surrounded by a lighter region<em>"penumbra"</em>.

Now, a solar protuberance or <u>prominence</u> is a large gaseous structure located on the surface of the Sun, often in a loop shape, which emerges from the surface of the Sun, the photosphere, and extends to reach the solar Corona. It is caused by disturbances in the Sun's magnetic field (hence it is<u> related to sunspots</u>) and, although most of the gas expelled returns to the surface, sometimes this can release particles that can reach the Earth.

In this context, a <u>solar flare</u> is a sudden, rapid and intense variation of brightness that is observed on the surface of the Sun due to an explosion of hot gases. It releases a large amount of energy in the form of electromagnetic radiation, energetic particles and mass in motion. These phenomena take place in active regions of the Sun, areas of strong magnetic field, and especially in sunspots with great magnetic complexity.

These eruptions are also known as Coronal Mass Ejections (CME), since they occur in the Sun's Corona layer; and can produce X-rays and gamma rays. However, <u>they are classified according to the peak of X-ray flow</u> by the letters A, B, C, M or X, where each letter (or class)  represents an increase in energy production 10 times greater than the previous one.

In summary:

<h2> When a sunspot is observed it is because in that region there was a coronal mass ejection, also known as a solar flare. This in turn is related to solar prominences. </h2>

7 0
3 years ago
List the two types of mechanical waves and define them.
Liula [17]
The two types of mechanical waves are longitudinal waves, and transversal waves.

7 0
3 years ago
Consider two soap bubbles with radius r1 and r2 (r1 &lt;r2) connected via a valve. What happens if we open the valve​
Sedaia [141]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is   \Delta  P _1 =10  J/m^3

The pressure difference of the second bubble is  \Delta  P _2 =20  J/m^3

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

    The  radius of the first bubble is  r_1 =  10 \ mm=0.01 \ m

      The radius of the second bubble is  r_2  =  5 \ mm  =  0.005 \ m

      The surface tension of the soap solution is  s =  25 \ mJ/m^2 = 25*10^{-3} J/m^2

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

         \Delta  P  =  \frac{4 s}{R}

Now the pressure difference for the first bubble is  mathematically evaluated as

        \Delta  P _1 =  \frac{4 s}{r_1}

substituting values  

       \Delta  P _1 =  \frac{4 *25 *10^{-3}}{0.01}

       \Delta  P _1 =10  J/m^3

Now the pressure difference for the second bubble is  mathematically evaluated as

        \Delta  P _2 =  \frac{4 s}{r_1}

       \Delta  P _2 =  \frac{4 *25 *10^{-3}}{0.005}

       \Delta  P _2 =20  J/m^3

3 0
3 years ago
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