All categories

3 years ago

1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of

Physics

1 answer:

nevsk [136]3 years ago

3 0

Answer:

$4.8 \cdot 10^4 N$

Explanation:

The force of gravity acting on the satellite is given by:

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

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

$r=2.4\cdot 10^6 m$

Substituting into the equation, we find:

$F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N$

Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).

You might be interested in

Both Slicing a tomato and a chemical change has a burning toast cannot be reversed. however, why is slicing a tomato still consi

olchik [2.2K]

Because the tomato did not chemically changed it was only physically altered.

Hope this helps! :D

6 0

3 years ago

Read 2 more answers

A 1 kg object sits on the earth’s surface. What is the force of gravity between the object and the earth? (mass of the earth = 5

snow_lady [41]

Answer:

9.81N

Explanation:

the force of attraction is given by

F=GmM/R²

where m is mass of the body

M is mass of the earth

R is radius of the earth

G is the universal gravitational constant(6.67x10-¹¹)

hence we substitute the values in the formula.

you can ask questions

4 0

2 years ago

Who was the last person to step on the moon?

Pavlova-9 [17]

Eugene Cernan was not the last, but he was the most recent.

5 0

2 years ago

Read 2 more answers

Please help..................

worty [1.4K]

Answer:

PART A

In a solid

The attractive forces keep the particles together tightly enough so that the particles do not move past each other. ... In the solid the particles vibrate in place. Liquid – In a liquid, particles will flow or glide over one another, but stay toward the bottom of the container.

In a liquid

Particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles.

A gas

The particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster.

PART B

The molecules are continually colliding with each other and with the walls of the container. When a molecule collides with the wall, they exert small force on the wall The pressure exerted by the gas is due to the sum of all these collision forces. The more particles that hit the walls, the higher the pressure.

Explanation:

GOOD LUCK!!! :)

7 0

2 years ago

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

2 years ago