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jeka57 [31]
3 years ago
6

Explain how changes in weather are caused by the interaction of air masses

Physics
1 answer:
zepelin [54]3 years ago
8 0
<span>When the lower air pressure gets colds it turns to be cold weather turns in to the high pressure like wise it would be changing of high pressure gets change in to the low pressure of it.So that there will be huge storm has been occurred and it will cause a rain.</span>
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20.
sveticcg [70]

Answer:

I believe the answer is A and D.

I am unsure of C.

4 0
3 years ago
Read 2 more answers
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
3 years ago
What happens to the rock structure during each of type of change?
EastWind [94]
Previous rocks melt and collide and to form igneous rocks.
Igneous rocks disintegrate due to weather disruptions and get carried away by water, where they form sedimentary rock strata by lithification.
Igneous and sedimentary change by heat and pressure to form metamorphic rocks.
Metamorphic rocks melt and become igneous rocks.
3 0
3 years ago
Which of the following is a subatomic particle?
mr Goodwill [35]

Answer:

neutron.

Explanation:

subatomic particles include,

neutron.

proton.

electron.

hope it helps. :)

4 0
3 years ago
Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal. What is the range of the ball?
just olya [345]

Answer:

the range or the ball is 48.81 m

Explanation:

given;

Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal.

find:

What is the range of the ball?

solution:

let Ф = 25°

Vo = 25 m/s

<u>consider x-motion using time of fight: x = Vox * t</u>

where x = R = range

t =<u> 2 Voy </u>

      g

R =<u> Vo² sin (2Ф)</u>

           g

plugin values into the formula:

R = <u>(25)² sin (2*25) </u>

               9.81

R = 48.81 m

therefore, the range or the ball is 48.81 m

4 0
4 years ago
Read 2 more answers
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