Answer:
The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Explanation:
From the question we are told that
The distance between earth and Retah is 
Here c is the peed of light with value 
The time taken to reach Retah from earth is 
The velocity of the spacecraft is mathematically evaluated as
substituting values
The time elapsed in the spacecraft’s frame is mathematically evaluated as
substituting value
![T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }](https://tex.z-dn.net/?f=T%20%20%3D%20%2090000%20%2A%20%20%5Csqrt%7B%201%20-%20%20%5Cfrac%7B%5B2.4%2A10%5E%7B8%7D%5D%5E2%7D%7B%5B3.0%2A10%5E%7B8%7D%5D%5E2%7D%20%7D)
=> 
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
1200
-------=171 miles per hour
7
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.
Answer:
q = 4.87 X 10^ -14 C
Explanation:
As d=0.350 mm
The ink drop will be accelerated by the electric field between the plates:
a = F/m
d = a(D0 / v)^2 / 2 ...... 1
a = qE/m ............... 2
Substituting 2 into 1:
d = (qE/m)(D0 / v)^2 / 2
q = 2mdv^2 / [E(D0)^2]
q = 2(1.00e-11 kg)(3.50e-4 m)(15.0 m/s)^2 / [(7.70e4 N/C)(2.05e-2 m)^2]
q = 4.87e-14 C
F = ma
Acceleration in this case is acceleration due to gravity so
a = 9.8 m/s^2
and mass = 0.57kg
So..
F = (0.57)(9.8)
F = 5.586 or 5.56 N
