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FinnZ [79.3K]
3 years ago
14

Luke Autbeloe drops an approximately 5.0 kg object (weight = 50.0 N) off the roof of his house into the swimming pool below. Upo

n encountering the pool, the object encounters a 50.0 N upward resistance force (assumed to be constant). Use this description to answer the following questions. (Down is usually considered a negative direction) a. Which one of the velocity-time graphs best describes the motion of the object? Why?

Physics
1 answer:
Murljashka [212]3 years ago
6 0

Answer:

Graph B represents correct graph for velocity time

Explanation:

Initially the object is dropped under the influence of constant force due to gravity

So here we can say its acceleration is constant due to gravity

hence here the velocity will increase downwards due to gravity and it is given as

v = 0 + (-10)t

now when it enters into the pool an upward force of same magnitude will act on it

so net force on it is

F = 50 - 50 = 0

so here we will have no acceleration and it will move with constant speed after that

so here correct graph is

Graph B

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0A: accelerating
AB: constant
BC: decelerating
CD:at rest
DE:accelerating
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hope this helps
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3 years ago
hree identical resistors are connected in series. When a certain potential difference is applied across the combination, the tot
pav-90 [236]

Answer:

The power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

Explanation:

Given;

three identical resistors connected in series

let the first resistor = R₁

let the second resistor = R₂

let the third resistor = R₃

Rt = R₁ + R₂ + R₃

Since the resistors are identical, thus, R₁ = R₂ = R₃ = R

Rt = 3R

Power is given as;

P = IV = V² / R

P = \frac{V^2}{R_t} = \frac{V^2}{3R} \\\\3P = \frac{V^2}{R} ------equation(i)

If the 3 identical resistor connection were changed to parallel, then the equivalent resistance in the circuit will be;

\frac{1}{R_t} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3} \\\\But, R_1 = R_2 = R_3\\\\\\frac{1}{R_t} = \frac{1}{R} +\frac{1}{R} + \frac{1}{R} \\\\\frac{1}{R_t} =\frac{3}{R} \\\\R_t = \frac{R}{3} \\\\P = \frac{V^2}{R_t} = \frac{3V^2}{R} \\\\P_{parallel} = \frac{3V^2}{R} ---------equation (ii)\\\\From \ equation \ (i), 3P_{series} = \frac{V^2}{R}, Substitute \ this \ into \ equation \ (ii)\\\\P = 3(\frac{V^2}{R} )\\\\P = 3(3P)\\\\P_{parallel} = 9P_{series}\\\\P_{parallel} = 9(45)\\\\

P_{parallel} = 405 \ W

Therefore, the power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

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