Answer:
v₂> v₃ velocity canoe is more than velocity fishing boat
Explanation:
For this exercise we must define a system consisting of the girl, Sally and the boat, in one case the canoe and in the other the fishing boat; for this system we can use moment conservation
Initial moment. Before the jump
p₀ = (M + m₂) v
Final moment. After the jump
= M v₁ - m₂ v₂
Where m and v are the masses and speed of the canoe
p₀ = p_{f}
(M + m₂) v = M v₁ - m₂ v₂
In the case of changing the canoe for the heaviest fishing boat, the final moment is
p_{f} = M v₁ - m₃ v₃
p₀ = p_{f}
(M + m₃) v = M v₁ - m₃ v₃
Since the canoe is stopped the speed v = 0, we write the speed of each boat
Canoe
0 = M v₁ - m₂ v₂
v₂ = M / m₂ v₁
Fishing boat
0 = M v₁ - m₃ v₃
v₃ = M / m₃ v₁
Since the masses of the fishing boat (m₃) is greater than the mass of the canoe (m₂) the speed of the fishing boat is less than the speed of the canoe, we can find the relationship between the two speeds
v₂ / v₃ = m₃ / m₂
Here you can see what v₂> v₃ velocity canoe is more than velocity fishing boat
Answer: If you meant 5 miles in 20 minutes than it’s 1 mile in 5 minutes
Explanation:
The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
brainly.com/question/14883923
Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.