how difficult is it to start a heavy lorry moving and to stop it moving? (choose one answer from the image provided)

A geosynchronous orbit can be defined as circular orbit which lies on the Earth's equatorial plane and follows the direction of the Earth's rotation in a period that's equal to the Earth's rotational period and thereby appearing motionless, at a fixed position in the sky relative to the ground observers.

We are given;

Radius of earth(R) = 6.37 x 10^(6) m

Mass of earth (Me) = 5.97 x 10^(24) kg

Gravitational constant = 6.67 × 10^(-11) m³/kg.s²

The earth has a rotational period of 24 hours per day. This gives in seconds

T = 24 × 60 × 60

T = 86400 s

Let's make the height of the orbit from Earth's surface to be h

Also, let ω be the uniform angular velocity in rad/s with which the satellite rotates in the geosynchronous orbit

Now, equating the centripetal force with the gravitational force gives us;

mω²(R + h) = G•Me•m/(R + h)²

m will cancel out. Also ω can be written as 2π/T

Thus,we now have;

(R + h) = ∛(G•Me•T²/(4π²))

Plugging in the relevant values, we have;

(R + h) = ∛(6.67 × 10^(-11) × 5.97 x 10^(24) × 86400²/(4π²))

(R + h) = 42227 Km

Since R = 6.37 x 10^(6)m = 6370 km

Thus;. h = 42227 - 6370 = 35857 km

3 0

3 years ago

A 0.0950-kg ball is thrown at 28.0 m/s toward a brick wall. 1) Determine the impulse that the wall imparts to the ball when it h

Wittaler [7]

1) $-5.32 kg m/s$

The impulse exerted by the wall on the ball is equal to the change in momentum of the ball:

$I=\Delta p = m (v-u)$

where

m = 0.0950 kg is the mass of the ball

v = -28.0 m/s is the final velocity of the ball (negative because it is away from the wall)

u = +28.0 m/s is the initial velocity of the ball (positive because it is towards the wall)

Substituting into the equation, we find

$I=(0.0950 kg)(-28.0 m/s-(+28.0 m/s))=-5.32 kg m/s$

2) $-3.76 kg m/s$

The problem is similar to before, but this time we must consider only the component of the initial and final velocities that are perpendicular to the wall. So we have:

$u_x = u sin 45^{\circ}=(+28.0 m/s)(sin 45^{\circ} C)=+19.8 m/s$ is the component of the initial velocity perpendicular to the wall