Answer:
If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.
Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:
PE = qV
where q =1.6 x 10^-19 C
and V = 1.00 V
Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:
PE* = 0,
KE* = 1/(2mv^2)
Now since
PE + KE = Total energy =PE* + KE*
Therefore,
qV + 0 = 0 + 1/2mv^2
Or
KE = qV = 1.6 10^-19 J
Answer:
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I hope it's helpful!
Answer:
Double the current
Explanation:
The energy delivered by the heater is related to the current by the following relation:
E= 
let R * t = k ( ∴ R and t both are constant)
so E= k 
Now let:
E2= k I₂^2
E2= 4E
⇒ k I₂^2= 4* k
Cancel same terms on both sides.
I₂^2= 4*
taking square-root on both sides.
√I₂^2 = √4* I^2
⇒I₂= 2I
If we double the current the energy delivered each minute be 4E.
Answer:
α = - 1.883 rev/min²
Explanation:
Given
ωin = 113 rev/min
ωfin = 0 rev/min
t = 1.0 h = 60 min
α = ?
we can use the following equation
ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t
⇒ α = (0 rev/min - 113 rev/min) / (60 min)
⇒ α = - 1.883 rev/min²
