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3 years ago

Describe the pattern of air circulation between an area of low pressure and high pressure

1 answer:

4 0

The sinking of cold air creates regions of high pressure and this air travels to lower pressure regions produced by the rising motion of warm air. This process results in air circulation.

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A world class sprinter is travelling with speed 12.0 m/s at the end of a 100 meter race. Suppose he decelerates at the rate of 2

Solnce55 [7]

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so he travel 36 m to stop

3 0

2 years ago

Even if all stars were the same distance from Earth, their absolute magnitude and

k0ka [10]

Answer: True

hope this helps!

4 0

2 years ago

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute

levacccp [35]

Explanation:

The given data is as follows.

Mass, m = 75 g

Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where, $m_{1}$ = mass of the projectile

$m_{2}$ = mass of block

v = velocity after the impact

Now, putting the given values into the above formula as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

$75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v$

= $\frac{45}{50.075}$

v = 0.898 m/s

Now, equation for energy is as follows.

E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

= 13500 J

Now, energy after the impact will be as follows.

E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

= 20.19 J

Therefore, energy lost will be calculated as follows.

$\Delta E$ = E E'

= (13500 - 20) J

= 13480 J

And, n = $\frac{\Delta E}{E}$

= $\frac{13480}{13500} \times 100$

= 99.85

= 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0

3 years ago

Read 2 more answers

Two satellites A and B orbit the Earth in the same plane. Their masses are 5 m and 7 m, respectively, and their radii 4 r and 7

Dmitry [639]

Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

$v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}$ ......(I)

The orbital speed of satellite B,

$v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}$ ......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

$\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}$

Put the value into the formula

$\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}$

$\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}$

Hence, The ratio of their orbital speeds are 5:4.

8 0

2 years ago

Name some elements that have a symbol that is entirely different from the spelling of the world

Alenkasestr [34]

Some elements that have a symbol entirely different from the spelling are..

1.Fe. Iron
2.Na.Sodium
3.K. Pottasium
4.Ag.Silver
5.Sn.Tin
6.Sb. Antimony
7,Pb.Lead

7 0

2 years ago