Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.
<u>Given the following data:</u>
- Gravitational constant = 6.67 × 10⁻¹¹ m/kg²
- Mass of Moon = 7.36 × 10²² kg
- Distance, r = 4.2 × 10⁶ m.
<h3>How to determine the speed of this satellite?</h3>
In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.
This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:
Fc = Fg
mv²/r = GmM/r²
<u>Where:</u>
- m is the mass of the satellite.
Making v the subject of formula, we have;
v = √(GM/r)
Substituting the given parameters into the formula, we have;
v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)
v = √(1,168,838.095)
v = 1,081.13 m/s.
Speed, v = 1.8 × 10³ m/s.
Read more on speed here: brainly.com/question/20162935
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Answer:
Yes it is possible
Explanation:
When two equal magnitude forces are acting on the rod in opposite direction
Then the net force on the system is always zero in that case
so we will have

now for the system net torque due to these forces is given by

here we know that
= distance of the forces from reference about which torque is measured
so here we can say that net force is zero on the system while torque is not zero
in all such case object will rotate about a fixed position with change angular speed
Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
The force due to air resistance is 256 N.
Explanation:
Given;
mass of the plane, m = 5 kg
applied force on the plane, Fa = 706 N
the net force on the plane, ∑F= 450 N
Let the force due to air resistance = Fr
The net force on the plane is given as;
Net force = applied force - force due to air resistance
∑F = Fa - Fr
Fr = Fa - ∑F
Fr = 706 - 450
Fr = 256 N.
Therefore, the force due to air resistance is 256 N.