10. Prepare summary notes on the topics of:
1 answer:
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Answer:38.675 m/s
Explanation:
t=4.2 s
Solving in x direction
ucos25=25
u=27.584 m/s
Initial velocity in vertical direction usin25
Let h be the height of Rooftop
h=37.47 m
Therefore final vertical velocity is
=29.51 m/s
Final Resultant velocity
Answer:
A) and B) are correct.
Explanation:
Let's take a look at the attached picture. Now
The total voltage across both capacitors is the same as the sum of the voltage from each device, that statement is true for any electrical device connected in series. So a) is TRUE
The equivalent capacitance is going to be:
And that value can be mathematically proven that is always less than any of the values of each capacitor. So b is TRUE
And through both capacitors flow the same current, but the amount of charge depends on the value of the capacitors, so only could be the same if the capacitors are the same value. Otherwise, don't. C) not always, so FALSE
Consider the travel of the speedboat from A to B and back.
Refer to the first figure.
The boat travels at 5 m/s relative to the water, and the downstream speed of the water is 0.5 m/s.
Therefore,
V₁=5 m/s, u = 0.5 m/s, sinθ = 0.5/5 = 0.1 => θ = arcsin(0.1) = 5.74°.
The boat should travel upstream at 5 m/s, at an angle of 5.74°.
Similarly, return speed from B to A is 5 m/s, at 5.74° upstream.
The horizontal component of velocity from A to B (or vice versa) is
5cos(5.74°) = 4.975 m/s.
The time required to travel 1000 m is
1000/4.975 = 202.02 = 3.367 min.
The time for the return trip is
t₁ = 2*3.367 = 6.734 min.
Consider travel from C to D and back, as shown in the second figure.
The resultant velocity upstream from C to D is
V₁ = 5 - 0.5 = 4.5 m/s
The time required to travel 1000 m fromC to D is
1000/4.5 = 222.22 s
The resultant velocity downstream from D to C is
V₂ = 5 + 0.5 = 5.5 m/s
The time required to travel fro D to C is
1000/5.5 = 181.82 s
Total time for the return trip between C and D is
t₂ = 222.22 + 181.82 = 404.04 s = 6.734 min
Answer:
The first boat should travel upstream at an angle of 5.74°. The time for the return trip between A and B is t₁ = 6.734 min or 404.04 s.
The second boat travels upstream against the current, and downstream with the current. The time for the return trip between C and D is t₂ = 6.734 min or 404.04 s.
Refer to the first figure.
The boat travels at 5 m/s relative to the water, and the downstream speed of the water is 0.5 m/s.
Therefore,
V₁=5 m/s, u = 0.5 m/s, sinθ = 0.5/5 = 0.1 => θ = arcsin(0.1) = 5.74°.
The boat should travel upstream at 5 m/s, at an angle of 5.74°.
Similarly, return speed from B to A is 5 m/s, at 5.74° upstream.
The horizontal component of velocity from A to B (or vice versa) is
5cos(5.74°) = 4.975 m/s.
The time required to travel 1000 m is
1000/4.975 = 202.02 = 3.367 min.
The time for the return trip is
t₁ = 2*3.367 = 6.734 min.
Consider travel from C to D and back, as shown in the second figure.
The resultant velocity upstream from C to D is
V₁ = 5 - 0.5 = 4.5 m/s
The time required to travel 1000 m fromC to D is
1000/4.5 = 222.22 s
The resultant velocity downstream from D to C is
V₂ = 5 + 0.5 = 5.5 m/s
The time required to travel fro D to C is
1000/5.5 = 181.82 s
Total time for the return trip between C and D is
t₂ = 222.22 + 181.82 = 404.04 s = 6.734 min
Answer:
The first boat should travel upstream at an angle of 5.74°. The time for the return trip between A and B is t₁ = 6.734 min or 404.04 s.
The second boat travels upstream against the current, and downstream with the current. The time for the return trip between C and D is t₂ = 6.734 min or 404.04 s.