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choli [55]
4 years ago
14

A low cloud that blankets the sky and often generates precipitation is called a(n) ____

Physics
2 answers:
DaniilM [7]4 years ago
4 0
The answer is Nimbostratus Cloud.

Hope this helps!! ;))
Have a grat day!! <3
denpristay [2]4 years ago
3 0
The answer is, "N<span>imbostratus cloud". </span>
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A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?
allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

7 0
3 years ago
Nvm figured it out 1!!!1!3!2!1!3!
Natali [406]

Answer:

that's great, good job!

6 0
3 years ago
A coil spring stretches by 2.50 cm when a mass of 750 g is suspended from it. (a) Find the force constant of the spring. (b) How
weqwewe [10]

Answer:

(a) 294 N/m

(b) 0.027 m or 2.7 cm

Explanation:

(a) From Hook's law,

F = ke......................... Equation 1.

make k the subject of the equation

k = F/e..................... Equation 2

Where F = force applied on the spring, k = spring constant, e = extension.

From the question, the force applied to the spring is the weight of the mass.

F = mg = 0.75(9.8) = 7.35 N

Also, e = 2.5 cm = 0.025 m

Substitute these values into equation 2

k = 7.35/0.025

k = 294 N/m.

(b) If a mass of 800 g is suspended,

F = ke

e = F/k................... Equation 3

F = mg = 0.8(9.8) = 7.84 N,

k = 294 N/m.

Substitute these values into equation 3

e = 7.84/294

e = 0.027 m

e = 2.7 cm

7 0
3 years ago
What is a dependent or responding variable? Can someone help me?
Svetllana [295]

They are a variable that changes as a result of the changes in the manipulated variable

8 0
3 years ago
A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide
-BARSIC- [3]

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

8 0
4 years ago
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