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3 years ago

The air in a cool region underneath cloud cover will have _____ a region with no cloud cover.

Physics

2 answers:

Slav-nsk [51]3 years ago

7 0

<u>Answer:</u>

The air is a <u>homogeneous mixture of gases</u> that make up the Earth's atmosphere, which remain around the planet by the attraction force of gravity.

Its properties changes with the altitude, the pressure, the temperature and the density. For example, there is a relation that is <u>inversely proportional</u> between the temperature and the air density.

<h2>When the temperature decreases the density increases. </h2>

Therefore, the air in a cool region of the atmosphere will be denser than the air in a hot region of the atmosphere.

In other words:

<h2>The air in a cool region underneath cloud cover will have <em><u>a higher density than</u></em> a region with no cloud cover </h2><h2> </h2>

Aleksandr-060686 [28]3 years ago

5 0

The air in a cool region underneath cloud cover will have a higher density than a region with no cloud cover. A region with no cloud cover has warmer air than the surrounding air, it becomes less dense and begins to rise, drawing more air in underneath.

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How Many Negative (-) Electrons are there

makvit [3.9K]

Answer:

All electrons are negative(-) charged

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3 years ago

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

$V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s$

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

Therefore, the final position made with the vertical is 2.77 m.

3 0

3 years ago

How much work is done on a 75 newton bowling ball when you carry it horizontally across a 10 meter room

svlad2 [7]

F = force applied to hold the weight of the bowling ball = weight of the bowling ball = 75 N

d = distance through which the bowling ball is moved horizontally = 10 meter

θ = angle between the force in vertically upward direction and displacement in horizontal direction = 90

W = work done on the bowling ball

work done on the bowling ball is given as

W = F d Cosθ

inserting the values

W = (75) (10) Cos90

W = (75) (10) (0)

W = 0 J

6 0

3 years ago

An object has a kinetic energy of 175 J and a momentum of magnitude 25.0 kg m/s. Find the

DedPeter [7]

Answer:14 m/s

Explanation:

Kinetic energy(ke)=175J

Momentum(M)=25kgm/s

Speed=v

Mass=m

Ke=(m x v x v)/2

175=(mv^2)/2

Cross multiply

175 x 2=mv^2

350=mv^2

Momentum=mass x velocity

25=mv

m=25/v

Substitute m=25/v in 350=mv^2

350=25/v x v^2

350=25v^2/v

v^2/v=v

350=25v

v=350/25

v=14 m/s

5 0

3 years ago

To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1

dedylja [7]

Answer:

$2.2\times 10^{-5} s$

Explanation:

We are given that

The wavelength of sound wave= $\lambda=3.3 cm=3.3\times 10^{-2}m/s$

1 cm/s= $10^{-2}m/s$

Speed of sound wave,v= $1522 m/s$

We have to find the period of the wave.

We know that

Frequency= $\nu=\frac{v}{\lambda}$

Using the formula

Frequency = $\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4}$ Hz

Time period= $\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s$

Using identity: $\frac{a^x}{a^y}=a^{x-y}$

Hence, the time period of the wave= $2.2\times 10^{-5} s$

4 0

3 years ago