A force of 200N acts between two objects at a certain distance apart .the value of the force when the distance is halved is?

Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and

Anna11 [10]

Kepler’s third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496 × 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/( 86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The closest answer is 1.99 × 10^30

(it may vary a little with rounding – the difference is less than 1%)

5 0

3 years ago

Read 2 more answers

An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.

Hitman42 [59]

Answer:

(a) max. height = 3.641 m

(b) flight time = 1.723 s

(c) horizontal range = 31.235 m

(d) impact velocity = 20 m/s

Above values have been given to third decimal. Adjust significant figures to suit accuracy required.

Explanation:

This problem requires the use of kinematics equations

v1^2-v0^2=2aS .............(1)

v1.t + at^2/2 = S ............(2)

where

v0=initial velocity

v1=final velocity

a=acceleration

S=distance travelled

SI units and degrees will be used throughout

Let

theta = angle of elevation = 25 degrees above horizontal

v=initial velocity at 25 degrees elevation in m/s

a = g = -9.81 = acceleration due to gravity (downwards)

(a) Maximum height

Consider vertical direction,

v0 = v sin(theta) = 8.452 m/s

To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,

solve for S in equation (1)

v1^2 - v0^2 = 2aS

S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s

(b) total flight time

We solve for the time t when the vertical height of the object is AGAIN = 0.

Using equation (2) for vertical direction,

v0*t + at^2/2 = S substitute values

8.452*t + (-9.81)t^2 = 3.641

Solve for t in the above quadratic equation to get t=0, or t=1.723 s.

So time for the flight = 1.723 s

(c) Horiontal range

We know the horizontal velocity is constant (neglect air resistance) at

vh = v*cos(theta) = 25*cos(25) = 18.126 m/s

Time of flight = 1.723 s

Horizontal range = 18.126 m/s * 1.723 s = 31.235 m

(d) Magnitude of object on hitting ground, Vfinal

By symmetry of the trajectory, Vfinal = v = 20, or

Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s

7 0

3 years ago

This is a table below

astra-53 [7]

Venus moves faster than earth because it move with a higher speed as compared to earth.

<h3>Is Venus move faster than earth?</h3>

Venus move with the speed of 35.02 km/s while on the other hand, the earth moves with the speed of 29.78 km/s. So due to more speed of Venus, the Venus completes its revolution earlier than earth.

So we can conclude that Venus moves faster than earth because it move with a higher speed as compared to earth.

Learn more about Venus here: brainly.com/question/2323914

#SPJ1

5 0

2 years ago

PLEASE HELP!!!!

weqwewe [10]

The answer is A. <span>Some work input is used to overcome friction. </span>

6 0

3 years ago

What is the initial vertical velocity?

Marat540 [252]

Answer:

Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends 3.79 s in the air.The initial vertical velocity is the vertical component of the initial velocity: v 0 y = v 0 sin θ 0 = ( 30.0 m / s ) sin 45 ° = 21.2 m / s .

3 0

2 years ago