R=√20²+40²+2.20.40 cos 50
R=55 N
R/sin β = F/sin α
55/sin 50 = 40/sin α
α = 33°
Mass of the car = 1200 kg
Mass of the truck = 2100 kg
Total mass of car and truck = 2100 + 1200 = 3300 kg
Since, the car pushes the truck. Hence, they will move together and will have same acceleration.
Let the acceleration be a.
According to Newton's second law:
F(net) = ma
F = 4500 N
4500 = 3300 × a
a = 1.36 m/s^2
Let the force applied by the car on truck be F.
F = F(net) on the truck
F = ma
F = 2100 × 1.36
F = 2856 N
Hence, the force applied by the car on the truck is 2856 N
This problem can be solved based on the rule of energy conservation, as the energy of the photon covers both the energy needed to overcome the binding energy as well as the energy of ejection.
The rule can be written as follows:
energy of photon = binding energy + kinetic energy of ejectection
(hc) / lambda = E + 0.5 x m x v^2 where:
h is plank's constant = 6.63 x 10^-34 m^2 kg / s
c is the speed of light = 3 x 10^8 m/sec
lambda is the wavelength = 310 nm
E is the required binding energy
m is the mass of photon = 9.11 x 10^-31 kg
v is the velocity = 3.45 x 10^5 m/s
So, as you can see, all the parameters in the equation are given except for E. Substitute to get the required E as follows:
(6.63x10^-34x3x10^8)/(310x10^-9) = E + 0.5(9.11 x 10^-31)(3.45x10^5)^2
E = 6.41 x 10^-16 joule
To get the E in ev, just divide the value in joules by 1.6 x 10^-19
E = 4.009 ev
Your answer is B :) because well the conditions are harsh.
Answer:
a) v = 141.9 m/s
b) v = 317.4 miles/h
Explanation:
a) How fast was he moving in meters per second?
Hence, the jet ski is moving at 141.9 meters per second.
b) How fast was he moving in miles per hour?
Therefore, the jet ski is moving at 317.4 miles per hour.
I hope it helps you!