Answer:
The asteroid's acceleration at this point is
Explanation:
The equation that governs the trajectory of asteroid is given by :

The velocity of asteroid is given by :

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0
So,
Acceleration,
Put t = 0.971 s

So, the asteroid's acceleration at this point is
and it is decelerating.
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
Answer:
KE = 10530 J or 10.53 KJ
Explanation:
The formula for kinetic energy is KE = 1/2 mv^2
Let's apply the formula:
KE = 1/2 mv^2
KE = 1/2 (65kg) (18m/s)^2
KE = 10530 J or 10.53 KJ
Answer:
51.94 ft/s²
257.63 ft/s
Explanation:
t = Time taken = 4 s
u = Initial velocity = 34 mi/h
v = Final velocity
s = Displacement = 615 ft
a = Acceleration
Converting velocity to ft/s

Equation of motion

Acceleration is 51.94 ft/s²

Final velocity at this time is 257.63 ft/s
Answer:
a = -1 m/s^2
Explanation:
Vi = 75 m/s
Vf = 25 m/s
t = 50 s
Plug those values into the following equation:
Vf = Vi + at
25 = 75 + 50a
---> a = -1 m/s^2