The answer is B. Bye because B those study speed.
Answer:Technician A
Explanation:
Technician A statement is correct as
The battery is required to start the vehicle which, in effect, rotates the alternator at sufficient speed to keep the battery charged. This means if the battery is low it is not possible to start the vehicle and thus we are unable to test the alternator.
That is the battery is pre-requisite to test the alternator. So the battery must be at least a 75 % charge to test the alternator.
Answer:
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Explanation:
Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.
Substituting the values of the variables into the equation, we have
ΔV = V₂ - V₁.
ΔV = 175 V - 33 V.
ΔV = 142 V
The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.
So, substituting the values of the variables into the equation, we have
ΔU = eΔV
ΔU = eΔV
ΔU = -1.602 × 10⁻¹⁹ C × 142 V
ΔU = -227.484 × 10⁻¹⁹ J
ΔU = -2.27484 × 10⁻²¹ J
ΔU ≅ -2.275 × 10⁻²¹ J
So, the required equation for the electric potential energy change is
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
