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borishaifa [10]
3 years ago
13

Suppose earth's mass increased but earth's diameter didn't change. Describe how the gravitational force between Earth and the ob

ject on its surface would change.
Physics
1 answer:
zheka24 [161]3 years ago
7 0

Answer:

It increases proportionally

Explanation:

The gravitational force between the Earth and an object on its surface is given by

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the Earth's radius

In this problem, the Earth's mass is increased, while the diameter (and therefore, the radius) doesn't change. From the equation, we see that the gravitational force is directly proportional to the Earth's mass: therefore, if the mass is increased, the force will increase as well by the same proportion (for example, if the mass is doubled, the force will double as well)

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Why is the warm air rising while the cold air is sinking
UNO [17]
Because cold air tends to more dense, and it's therefore heavier, and it sinks.
Hot air though is hot with no density so it's light and rises.
6 0
3 years ago
Read 2 more answers
hot-air balloon is ascending at the rate of 14 m/s and is 84 m above the ground when a package is dropped over the side. (a) How
timurjin [86]

Answer:

a) t = 4.14 s

b) Speed with which it hits the ground = 40.58 m/s

Explanation:

Using the equations of motion,

g = 9.8 m/s², y = H = 84 m,

Initial velocity, u = 0 m/s,

final velocity, v = ?

Total Time of fall, t = ?

a) y = ut + gt²/2

84 = 0 + 9.8t²/2

4.9t² = 84

t² = 84/4.9

t = 4.14 s

b) v = u + gt

v = 0 + (9.8 × 4.14)

v = 40.58 m/s

4 0
3 years ago
Read 2 more answers
A car covers 400 km in an hour towards west .calculate the velocity​
crimeas [40]

Answer:

-400km/hr

Explanation:

Velocity=displacement/time

=400/1

=400Km/hr

=-400km/hr (because west direction)

7 0
3 years ago
A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th
Evgen [1.6K]

Answer:

\dfrac{d\theta}{dt}=0.038\ rad/s

Explanation:

Given that

\dfrac{dx}{dt}= -1\ m/s

From the diagram

tan\theta=\dfrac{21}{x}

By differentiating with time t

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

When x= 10 m

tan\theta=\dfrac{21}{10}

θ = 64.53°

Now by putting the value in equation

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)

\dfrac{d\theta}{dt}=0.038\ rad/s

Therefore rate of change in the angle is 0.038\ rad/s

8 0
3 years ago
What is the value of the activation energy of the uncatalyzed reaction in reverse?
alexgriva [62]
It would be: Activation Energy = 300 KJ

Hope this helps!
5 0
3 years ago
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