Answer:
v = √2G 
/ R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1 / R = - G m1 / R
v² = 2G (1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G / R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1 / R
Em = - G m1 / R
R = int ⇒ Em = 0
In technical terms, every coil of wire increases the "magnetic flux density" (strength) of your magnet.
So it's A (magnetic field increase)
In my opinion the answer is B. Variation
Pressure = force ( in newtons ) / area ( in m^2 )
pressure put
= 30 000 N / 0.75 m^2
= 40 000 Pa
Answer:
Option C. 30 m
Explanation:
From the graph given in the question above,
At t = 1 s,
The displacement of the car is 10 m
At t = 4 s
The displacement of the car is 40 m
Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:
Displacement at t = 1 s (d1) = 10 m
Displacement at t= 4 s (d2) = 40
Displacement between t = 1 and t = 4 (ΔD) =?
ΔD = d2 – d1
ΔD = 40 – 10
ΔD = 30 m.
Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.
