A marathon runner completes a 42.188–km course in 2 h, 36 min, and 12 s. there is an uncertainty of 23 m in the distance travele
1 answer:
The percentage uncertainty in the distance is <u>0.054%</u>.
Uncertainty in a measurement measures the deviation of the measured values from the true value.
The distance d can be expressed in the form <em>d+/-Δd, </em>where <em>Δd i</em>s the absolute uncertainty in the measurement of distance.
The percentage uncertainty is given by,
The uncertainty in the measurement of distance is 23 m. Express the uncertainty in km.
Calculate the percentage uncertainty in the distance.
Thus, the percentage uncertainty in the measurement of distance is <u>0.054%</u>
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Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
Answer:
Approximately . (Assuming that the drag on this ball is negligible, and that
.)
Explanation:
Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:
- Horizontal: no acceleration, velocity is constant (at
is constant throughout the descent.)
- Vertical: constant downward acceleration at
, starting at
.
The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: . Combine these two quantities to find the duration of this descent:
.
In other words, the ball in this question start at a vertical velocity of , accelerated downwards at
, and reached the ground after
.
Apply the SUVAT equation to find the vertical displacement of this ball.
.
In other words, the ball is below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be
.