Answer:
a) 
b) 
c) 
d) 
e) 
Explanation:
At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.
(1)
Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.
Let's solve (1) for β.
We can write the mass of a proton in MeV/c².
Now we can calculate the speed in each stage.
a) Cockcroft-Walton (750 keV)
b) Linac (400 MeV)
c) Booster (8 GeV)
d) Main ring or injector (150 Gev)
e) Tevatron (1 TeV)
Have a nice day!
The answer is C
Hope this helps!
Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :
We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,
On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.
<span>The number of oscillations that a wave completes per unit of time is called "frequency".
Frequency represents in fact how many complete cycles of the wave are done in a single unit of time. It can be written also as
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Answer:
The change in the mechanical energy of the projectile is 43,750 J
Explanation:
Given;
mass of the projectile, m = 5 kg
initial velocity of the projectile, u = 200 m/s
final velocity of the projectile, v = 150 m/s
The change in mechanical energy is calculated from the principle of conservation of energy;
ΔP.E = ΔK.E
The change in potential energy is zero (0)
0 = ΔK.E
ΔK.E = K.E₁ - K.E₂
ΔK.E = ¹/₂mu² - ¹/₂mv²
ΔK.E = ¹/₂m(u² - v²)
ΔK.E = ¹/₂ x 5(200² - 150²)
ΔK.E = 43,750 J
Therefore, the change in the mechanical energy of the projectile is 43,750 J
