Answer:
a. 2v₀/a b. 2v₀/a
Explanation:
a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.
Since the dragster starts from rest with an acceleration, a, using
s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster
s' = 0t + 1/2at²
s' = 1/2at²
Since the distance moved by me and the dragster must be the same,
s = s'
v₀t. = 1/2at²
v₀t. - 1/2at² = 0
t(v₀ - 1/2at) = 0
t= 0 or v₀ - 1/2at = 0
t= 0 or v₀ = 1/2at
t= 0 or t = 2v₀/a
So the maximum time tmax = 2v₀/a
b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)
= 2v₀/a
Answer:
0.03167 m
1.52 m
Explanation:
x = Compression of net
h = Height of jump
g = Acceleration due to gravity = 9.81 m/s²
The potential energy and the kinetic energy of the system is conserved

The spring constant of the net is 20130.76 N
From Hooke's Law

The net would strech 0.03167 m
If h = 35 m
From energy conservation

Solving the above equation we get

The compression of the net is 1.52 m
Answer:
, 
Explanation:
The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

where
k is the Coulomb constant
e is the magnitude of the charge of the electron
e is the magnitude of the charge of the proton in the nucleus
r is the distance between the electron and the nucleus
v is the speed of the electron
is the mass of the electron
Solving for v, we find

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

while the electron mass is

and the charge is

Substituting into the formula, we find

Answer:
Explanation:
First we calculate the energy of the photon
E=(Planck constant × speed of light in vacuum)÷ wave length
E=
Next we find the total energy per second
total energy= 
Next we calculate the number the photon per second
= total energy ÷ energy of 1 photon
= 