Answer:
a) 
b)
c) 
d) Treat the humans as though they were points or uniform-density spheres.
Explanation:
Given:
- mass of Mars,
- radius of the Mars,
- mass of human,
a)
Gravitation force exerted by the Mars on the human body:
where:
= gravitational constant
b)
The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.
c)
When a similar person of the same mass is standing at a distance of 4 meters:
d)
The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.
- Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
- Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
The answer is C. Final position minus initial position.
Answer:
False
Explanation:
When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.
As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.
ωd = ω₀√(1 - ζ)
Where ζ is called damping ratio.
For small value of ζ
ωd ≈ ω₀
Explanation
(m) is measured in kilograms (kg)
<h2>(F) is measured in newtons (N)</h2>
<h3>acceleration (a) is measured in metres per second squared (m/s²)</h3>
