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3 years ago

Does a front always mean one air mass will be displaced by another? When is there no displacement?

Physics

1 answer:

tangare [24]3 years ago

4 0

Answer:

Not necessarily, if the air masses are not strong enough, a stationary front is caused

Explanation:

A front is a block of separation between two air masses at different temperatures. They can be classified as cold, warm, stationary and occluded fronts. When a front is formed, a collision occurs between the two masses, generating a very dynamic activity, with electrical storms, gusts of wind and downpours.

A stationary front is one where neither of the two air masses are strong enough to replace the other. On such a front can generate atmospheric conditions such as prolonged rainfall and high cloudiness.

After several days, the stationary fronts may dissipate or become a cold or warm front. .

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You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes. That is half of your blood

Zina [86]

Answer:

8 times

Explanation:

Given that You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes.

That means the heart will pump 10 quarts in 2 minutes.

That is half of your blood volume per minute.

If during exercise it can pump 40 quarts per minute, that is, 80 quarts in 2 minutes.

To know how many times does all of your blood complete the cycle around your body during exercise, you must divide 80 quarts by 10 quarts. That is,

80 / 10 = 8

Therefore, your blood complete the cycle around your body 8 times during the exercise.

3 0

3 years ago

On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced

Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

$v =f\lambda$

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

$\mu = \frac{m}{l}$

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

$\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m$

Now, for the tension we use the formula:

$v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T$

T = 0.015 N

(b)

The mass to be hung is:

$T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\$

M = 1.53 x 10⁻³ kg = 1.53 g

4 0

3 years ago

A 7 kg ball is moving at a constant speed of 5 m/s. A force of 300 N is applied to the ball for 4 s. The new speed of the ball i

olasank [31]

Answer:

The new speed of the ball is 176.43 m/s

Explanation:

Given;

mass of the ball, m = 7 kg

initial speed of the ball, u = 5 m/s

applied force, F = 300 N

time of force action on the ball, t = 4 s

Apply Newton's second law of motion;

$F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v = \frac{Ft}{m} + u$

where;

v is new speed of the ball

$v = \frac{Ft}{m} + u\\\\v =\frac{300*4}{7} + 5\\\\v = 176.43 \ m/s$

Therefore, the new speed of the ball is 176.43 m/s

8 0

3 years ago

Someone please help its a simple power problem.

SOVA2 [1]

Well 200 doubled or (x2)=400 if that’s what it means

7 0

2 years ago

Which of the following affect friction (you can choose more than 1)

Anastaziya [24]

I think it would be WEIGHT and ROUGHNESS OF SURFACE.

4 0

3 years ago

Read 2 more answers