Ask question

Ask question

All categories

3 years ago

9

A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height of 3200 km fr

om the earth's surface? (Radius of the earth is 6400 km)

1 answer:

natali 33 [55]3 years ago

4 0

Answer:

you are kind of better than I am I don't read much but with brainly I will be intelligent

You might be interested in

A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.

Vlad1618 [11]

Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s

Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²

Answer: 44.7 m/s²

6 0

3 years ago

action and reaction are equal in magnitude and opposite in direction then why do they not balance each other.

NeX [460]

Answer:

Newton’s third law of motion states that every action has an equal and opposite reaction. This indicates that forces always act in pairs. Reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don’t cancel each other out.

7 0

3 years ago

A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide

-BARSIC- [3]

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

8 0

4 years ago

PLEASE HELP FOR BRAINLIEST ANSWER!

Alex787 [66]

I believe the answer is the fourth one, hope this helps

3 0

4 years ago

Read 2 more answers

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with <u>uniform circular motion</u>, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the <u>centripetal acceleration</u>

$V$ is the<u> tangential speed</u>

$r=12 m$ is the <u>radius</u> of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

<u />

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago