(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.
(b) The speed of the proton is determined as 3.1 x 10⁹ m/s.
<h3>
Momentum of the proton</h3>
The momentum of the proton is calculated as follows;
K.E = ¹/₂mv²
where;
- m is mass of proton = 1.67 x 10⁻²⁷ kg
- v is speed of the proton = ?
<h3>Speed of the proton</h3>
v² = 2K.E/m
v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)
v² = 9.6 x 10¹⁸
v = 3.1 x 10⁹ m/s
<h3>Momentum of the proton</h3>
P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s
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Answer:
d = 10.2 m
Explanation:
When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

where,
m = mass of car
v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)
v = 10 m/s
F = force on the car in direction of inclination = W Sin θ
W = weight of car = mg
θ = Angle of inclinition = 30°
d = distance covered up the ramp = ?
Therefore,

<u>d = 10.2 m</u>
control is constant for given set of readings.
set indep var
observe dep var
Given: Mass m = 0.50 Kg; Force = Weight = mg F = (0.50 Kg)(9.8 m/s²)
F = 4.9 N
Displacement x = 3.0 cm convert to meter x = 0.03 m
Required: Spring constant k = "
Formula: F = kx
k = F/x
k = 4.9 N/0.03 m
k = 163.33 N/m