A capacitor with an initial potential difference of 100 V isdischarged through a resistor when a switch between them is closed a
1 answer:
Answer:
(a). The time constant of the circuit is 2.17.
(b). The potential difference across the capacitor at t=17.0 s is 0.0396 V.
Explanation:
Given that,
Initial potential difference = 100 V
Potential difference across the capacitor = 1.00 V
(a). We need to calculate the time constant of the circuit
Using formula of potential difference
Put the value into the formula
On taking ln
(b). We need to calculate the potential difference across the capacitor at t=17.0 s
Using formula again
Hence, (a). The time constant of the circuit is 2.17.
(b). The potential difference across the capacitor at t=17.0 s is 0.0396 V.
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Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
- Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
- Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
- vₓ₀ = v * cos θ (1)
- where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
- Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:
- Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:
- ⇒θ = cos⁻¹ (0.526) = 58.3º (4)
b)
- At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
- Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
- vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)
c)
- At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)
d)
- Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:
- Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
- v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
- Replacing (7) in (6), we get:
e)
- When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
- The horizontal component, since it keeps constant, is just v₀x:
- v₀ₓ = 13.7 m/s
- The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:
- Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:
- Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows: