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morpeh [17]
3 years ago
9

The mass of earth 6*10^24kg and radius is 400kg Now find the value of acceleration due to gravity when a object is 3600km from t

he earth surface​
Physics
1 answer:
vlada-n [284]3 years ago
5 0
I don’t worry wewwwww it is a good time to get it done lol lol i don’t worry about it lol lol i lol
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A sphere is charged with electrons to
pishuonlain [190]
A sphere is charged with electrons to −9 × 10−6 C. The value given is the total charge of all the electrons present in the sphere. To calculate the number of electrons in the sphere, we divide the the total charge with the charge of one electron.

N = 9 × 10−6 C / 1.6 × 10−19 C
N = 5.6 x 10^13
3 0
3 years ago
Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second
ki77a [65]
To answer that question, we don't care what the highest and lowest
levels of the wave are, or how far apart they are.  We only need to be
able to identify the highest point on the wave, and keep track of how
often those pass by us.

You said it takes 4 seconds for a complete wave to pass by.
Through the sheer power of intellect, I'm able to take that information
and calculate that  1/4  of the wave passes by in 1 second.

There's your frequency . . .  1/4 per second, or  0.25 Hz.
6 0
3 years ago
A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication
koban [17]

Answer:

r = 4.24x10⁴ km.  

     

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0
3 years ago
A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0
3 years ago
Franny drew a diagram to compare images produced by concave and convex lenses.
yanalaym [24]

Answer:

virtual

Explanation:

4 0
3 years ago
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