Which is an example of precipitation?

A small planet having a radius of 1000 km exerts a gravitational force of 100 N on an object that is 500 km above its surface. I

Mars2501 [29]

<h2>Option C is the correct answer.</h2>

Explanation:

Gravitational force is given by

$F=\frac{GMm}{r^2}$

G=6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²

M = Mass of object 1

m = mass of object 2

r = Distance between objects.

Here only variable is r value.

In case 1

$100=\frac{GMm}{(1000+500)^2}\\\\GMm=100\times 1500^2$

In case 2

$F=\frac{GMm}{(1000+500+500)^2}\\\\F=\frac{GMm}{2000^2}\\\\F=\frac{100\times 1500^2}{2000^2}\\\\F=56.25N$

Option C is the correct answer.

7 0

3 years ago

A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be

nirvana33 [79]

Answer:

Work done, W = 19.6 J

Explanation:

It is given that,

Mass of the block, m = 5 kg

Speed of the block, v = 10 m/s

The coefficient of kinetic friction between the block and the rough section is 0.2

Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

$W=\mu_kmgd$

$W=0.2\times 5\times 9.8\times 2$

W = 19.6 J

So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

A loudspeaker, mounted on a tall pole, is engineered to emit 75% of its sound energy into the forward hemisphere, 25% toward the

Naily [24]

Answer:

"0.049 W" is the correct answer.

Explanation:

According to the given question,

$r = \sqrt{(3.5)^2+(2.5)^2}$

$=\sqrt{8.5}$

$SL=85$

As we know,

⇒ $SL=10 \ log(\frac{I}{I_o} )$

$85=10 \ log(\frac{I}{10^{-12}} )$

$I=3.162\times 1^{-4} \ W/m^2$

Now,

⇒ $P_{front} = I(2\pi r^2)$

$=(3.162\times 10^{-4})(2\pi\times 18.5)$

$=0.0368 \ W$

$=0.75 \ P$

or,

$=0.049 \ W$

7 0

3 years ago

If a 2000 N force is used to raise a piano at a rate of 10 m/s2, what is the mass of the piano?

saveliy_v [14]

Answer:

To solve this problem we will apply the principle of conservation of energy for which we have that the potential energy on a body, is equivalent to the work done on it at the given point. Therefore we will have the following equality

At the same time we know that work is equivalent to the Force applied over a given distance, so,

The potential energy is equivalent to the product between mass, gravity and height. Recall that the product of mass and gravity is equivalent to weight (The same given in the statement)

Equating,

Then,

Replacing,

Therefore the force needed to lift the piano is 600N

Explanation:

HOPE THIS HELPS!!!

3 0

3 years ago

Many physical properties, such as force and mass, cannot be measured directly. Rather, some other physical property is measured

Salsk061 [2.6K]

Answer:

u =0.269

Explanation:

To find the coefficient of friction we know the following formula

$F_{f} = uN$

Where

$F_{f}$ = Force of Friction

$u$ = Coefficient of Friction

$N$ = Normal Force

Thus we first find the Normal force (N). Remember that the Normal force is perpendicular to the surface, and is equal to the opposing component of Weight (W). Since the surface here is horizontal, then the Normal force will be equal to the Weight.

$N = mg\\ N = (2.12)(9.8)\\ N = 20.776 N$