What is sound reverberation?

If a satellite weighs 321 lb. on the earth's surface (R = 4,000 miles), how much does it weigh 12,000 miles above the surface? (

Sever21 [200]

The gravitational force between the Earth and the satellite (its "weight") is inversely proportional to the distance between the centers of both objects.

On the surface, their centers are separated by 1 Earth radius.

12,000 miles above the surface, they're separated by 4 Earth radiii.

(4/1) = 4

So after the move, the satellite's weight is (1/4²) = 1/16 of its surface weight.

(321 lb) / (16) = (20 and a hair) lb

The correct choice from the given list is " <em>>20 lb "</em> .

3 0

3 years ago

An ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Bond [772]

The y-component of the acceleration is $0.33 m/s^2$

Explanation:

The y-component of the acceleration is given by:

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

$u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s$

where

u = 2.25 m/s is the initial velocity

$\theta_1 = 50.0^{\circ}$ is the initial direction

$v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s$ , where

v = 4.65 m/s is the final velocity

$\theta_2 = 120.0^{\circ}$ is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

$a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2$

Learn more about acceleration:

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#LearnwithBrainly

3 0

3 years ago

A kangaroo has a maximum gravitational potential energy during one jump of 770 J

KengaRu [80]

Answer:

k = 44000 N/m

Explanation:

Given the following data;

Maximum gravitational potential energy = 770 J

Elastic potential energy = 14% of 770 J = 14/100 * 770 = 107.8 J

Extension, x = 42 - 35 = 7cm to meters = 7/100 = 0.07 m

To find the spring constant, k;

The elastic potential energy of an object is given by the formula;

$E.P.E = \frac {1}{2}kx^{2}$

Substituting into the equation, we have;

$107.8 = \frac {1}{2}*k*0.07^{2}$

$107.8 = \frac {1}{2}*k*0.0049$

Cross-multiplying, we have;

$215.6 = k*0.0049$

$k = \frac {215.6}{0.0049}$

k = 44000 N/m

5 0

3 years ago

710 mm= ______________cm<br>

love history [14]

Answer:

71 cm

Explanation:

Every 100 mm is equal to 10 cm. Hope this helps!

4 0

3 years ago

Read 2 more answers

The rock and meterstick balance at the 25-cm mark, as shown in the sketch. The meterstick has a mass of 1 kg. What must be the m

liraira [26]

Answer:

Explanation:

Check the diagram attached below for the diagram.

Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark

Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.

Moment = Force * perpendicular distance

The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction

Clockwise moment = 1kg * 25 = 25kg/cm

Anticlockwise moment = W * 25cm = 25W kg/cm

Equating both moments of forces

25W = 25

W = 25/23

W = 1 kg

The mass of the rock is also 1 kg

3 0

3 years ago