All categories

3 years ago

The centripetal force acting on a satellite in orbit

2 answers:

8 0

The centripetal force (of gravity) on a satellite in orbit is an
unbalanced force (because there's no equal force pulling
the satellite away from Earth), changes the direction of the
satellite (into a closed orbit instead of a straight line), and
always acts toward the center of whatever curve the satellite
happens to be on at the moment.

AysviL [449]3 years ago

3 0

Answer : (D) " all of the above "

Explanation :

The forces acting on a satellite are Earth's gravitational force and centripetal force. The properties of centripetal force acting on a satellite in orbit are as follows :

(1) It acts as an unbalanced force on the satellite. This is due to the reason that there is no same force present which pulls the satellite away from the earth.

(2) It changes the direction of the satellite. The satellite is revolving with a constant speed while the direction is kept on changing (in a circular motion).

(3) Centripetal force always acts towards the center.

So, the correct option is (D) " all of the above "

You might be interested in

Um corpo de massa igual a 2kg move-se com velocidade constante num plano horizontal sem atrito, conforme a figura. Em seguida, e

castortr0y [4]

Only answering to ask a question

4 0

3 years ago

An object is thrown upwards with the speed of 40.0 km/hr in 5 second, calculate the distance in millimeters (mm).

sertanlavr [38]

V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s

v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm

4 0

3 years ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,

$Q_A$ = $15\times 10^-6 C$

$Q_B=9\times10^-6C$

$d_A = 1+x cm$

$d_B=x cm$

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

$\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}$

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

3 0

1 year ago

David rowed a boat upstream for three miles and then returned to point he started from. The entire journey took four hours. The

frez [133]

Upstream speed = S - 1
Downstream speed = S + 1

Average speed = total distance / total time

Average speed = (S - 1) + (S + 1) / 2
= S

S = 6 miles / 4 hours
S = 1.5 miles per hour

4 0

3 years ago

Can life be found/ sustain on Mars or other planets? What conditions are required to support life on another planet?

Mariana [72]

No. Oxygen( an atmosphere to contain the oxygen),water ,sunlight(energy)

7 0

3 years ago