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2 years ago

Why does air blows from one place to another

Physics

2 answers:

Maslowich2 years ago

3 0

Air blows from one place to another because gases move from high-pressure areas to low-pressure areas

In simple words

it happens because of pressure differences.

mr Goodwill [35]2 years ago

3 0

Answer:

Gases moving from high-pressure places to low-pressure places. The bigger the difference of pressure the faster the air moves from both. Giving the movement of air we experience.

OR:

The high-pressure air will raise up to the low-pressure air and will come down causing the air will blow from one place to another.

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A ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum he

Andrew [12]

Answer:

10.2 metres

Explanation:

Given that a ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum height it will reach?

Solution

To get the maximum height, let us use the formula

V^2 = U^2 sin^2ø - 2gH

At maximum height V = 0

U^2 sin^2ø = 2gH

Substitute all the parameters into the formula

20^2 ( sin 45 )^2 = 2 × 9.8 × H

400 × 0.5 = 19.6 H

Make H the subject of formula

H = 200 / 19.6

H = 10.204 metres.

Therefore, the maximum height reached by the projected ball is 10.2 metres.

5 0

3 years ago

Under which conditions do both convex and concave mirrors form virtual images? State whether or not each can form a real image.I

natali 33 [55]

Concave mirrors can make a real image.when the object is away from the mirror it is called real image and when the object is closer to the mirror it is called virtual mirror.

8 0

3 years ago

Read 2 more answers

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle res

zhannawk [14.2K]

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

$w = \sqrt{\frac{k}{m}}$

$w = \sqrt{\frac{250}{0.5}}$

$w$ = 22.4 rad/s

$f$ = frequency

Angular frequency is also given as

$w$ = 2 π $f$

22.4 = 2 (3.14) f

$f$ = 3.6 Hz

$T$ = Time period

Time period is given as

$T = \frac{1}{f}$

$T = \frac{1}{3.6}$

$T$ = 0.28 sec

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

5 0

4 years ago

An Olympic diver springs off of a high dive that is 3 m above the surface of the water. When she lands in the water she is trave

klemol [59]

Answer: $u=4.51 m/s\ at\ angle\ of\ \theta =59.34^{\circ}$

Explanation:

Given

height of building h=3 m

Landing velocity of diver $v=8.90 m/s$ at an angle of $75^{\circ}$

Let u be the initial velocity of diver at an angle of \theta with horizontal

Since there is no acceleration in horizontal direction therefore horizontal component of velocity will remain same

$u\cos \theta =8.9\cos (75)$ ---- -----1

Considering Vertical motion

$v^2-u^2=2as$

here $v=8.9\sin (75)$

$u=u\sin \theta$

$s=3 m$

$a=9.8 m/s^2$

$(8.9\sin (75))^2-(u\sin \theta )^2=2\times 9.8\times 3$

$u\sin \theta =\sqrt{(8.9\sin 75)^2-(2\cdot 9.8\cdot 3)}$

$u\sin \theta =3.886$ ----------------2

Divide 2 and 1 we get

$\tan \theta =\frac{3.886}{8.9\cos (75)}$

$\tan \theta =1.687$

$\theta =59.34^{\circ}$

Thus $u\cos (59.34)=8.9\cos (75)$

$u=4.51 m/s$

7 0

3 years ago

Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e

NeTakaya

We have that for the Question "Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

$Q=\frac{E}{Nr^2}$

From the question we are told

Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the magnitude of charge moved</h3>

Generally the equation for the magnitude of charge moved, Q is mathematically given as

$Q=\frac{E}{Nr^2}$

Therefore

An expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e" it can be

$Q=\frac{E}{Nr^2}$

For more information on this visit

brainly.com/question/16517842

3 0

2 years ago

Why does air blows from one place to another​

Why does air blows from one place to another