Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
Answer:
A cyclotron is a type of particle accelerator.
Explanation:
Cyclotron frequency is the frequency of a charged particle moving perpendicular to the direction of a uniform magnetic field B, since that motion is always circular, the cyclotron frequency is given by equality of centripetal force and magnetic Lorentz force.
Answer:
1000 N
Explanation:
First, we need to find the deceleration of the running back, which is given by:
where
v = 0 is his final velocity
u = 5 m/s is his initial velocity
t = 0.5 s is the time taken
Substituting, we have
And now we can calculate the force exerted on the running back, by using Newton's second law:
so, the magnitude of the force is 1000 N.
Answer:
B) 18,000 feet MSL
Explanation:
There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.
Let us take east and north as the positive x and y-axes should the motion be plotted in a cartesian plane. Thus, the x value is 45 miles and the y value is 20. The tangent of an angle is equal to the ratio of y to x.
tanθ = y / x
Substituting,
tanθ = 20/45 = 0.44
The value of θ is 23.96°.
