<span>The land between two normal faults moves upward to form a
Answer:D</span><span>
fault-block mountain.</span>
Answer:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e

N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e

= 1250W
d) Final peak=
P= Ik/e
= 
P = 2.5 × 10⁷W
Answer:
Mass of the climber = 69.38 kg
Explanation:
Change in length

Load, P = m x 9.81 = 9.81m
Young's modulus, Y = 0.37 x 10¹⁰ N/m²
Area

Length, L = 15 m
ΔL = 5.1 cm = 0.051 m
Substituting
Mass of the climber = 69.38 kg
Answer:

Explanation:
From the question we are told that:
Acceleration 
Displacement 
Initial time 
Final Time 
Generally the equation for Velocity of 1.05 travel is mathematically given by
Using Newton's Law of Motion



Generally the equation for Distance traveled before stop is mathematically given by



Generally the equation for Distance to stop is mathematically given by
Since For this Final section
Final velocity 
Initial velocity 
Therefore
Using Newton's Law of Motion


Giving

Therefore



Generally the Total Distance Traveled is mathematically given by


