1) 4min = 4*60 sec = 240 sec
2) Distance = speed times time = 3 * 240 = 720 m
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = 
v²
v = √( 2K /
)
lets relate the cross-sectional area A of the beam to its diameter D;
A =
πD²
now, we substitute for v and A
n = I /
πeD² ×√( 2K /
)
n = 4I/π eD² × √(
/ 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
photons and convection - density differences makes bubbles of hot stuff float up. pretty sure
Answer:
B
Explanation:
I hope this is what you need
PLEASE MAKE ME BRAINLIEST
Answer:
<em>Air pockets.</em>
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Explanation:
Air pockets in the cooling system are bubbles of air trapped within the lines (hoses and pipes) of the cooling system. This air bubbles enter the cooling system usually during the process of filling the radiator coolant fluid (usually water), or replacing the water pump or the radiator hose during repairs or servicing of the cooling system. <em>The trapped air prevent pressure movement that is needed by the coolant to move the heat generated from the engine cylinder, resulting in heat build up</em>. The solution is to "bleed" the engine through the radiator lid or some air release valves.