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Eduardwww [97]
3 years ago
8

Arrange an 8-, 12-, and 16-Ω resistor in a combination that has a total resistance of 8.89 Ω pls with de work

Physics
1 answer:
kolezko [41]3 years ago
8 0


20 ohms in parallel with 16 ohm= 8.89

20x16/20+16. Product over sum


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Four forces are exerted on an object as shown below: A square is shown with an arrow pointing up labeled 3 N, an arrow pointing
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Answer:— A square labeled Object B is shown with an arrow pointing up labeled 3 N, an arrow pointing right labeled 2 N, an arrow pointing down labeled 3 N and an arrow pointing left labeled 1 N. In which direction will the objects move? Object A will move down and Object B will move up.

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A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/
Furkat [3]

Answer:

1) 23.45 rad/s²

2) 2.7 m/s²

3) t= 1.6 s

4) x ≈ 11 m

5) vfinal = 4.45 m/s

6) KErot = 16.2 J

    KEtran = 41 J

    KErot < KEtran

Explanation:

Step 1: Data given

mass bowling ball = 4.1 kg

radius = 0.117 meter

initial speed = 8.9 m/s

1) What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?

α = a / r = 2.774 m/s² / 0.117m = 23.45 rad/s²

2)What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?

a = µ*g = 0.28 * 9.8m/s² = 2.744 m/s² ≈ 2.7 m/s²

3) How long does it take the bowling ball to begin rolling without slipping?

This begins when ω = v / r

with

⇒ ω = α*t = 23.45 rad/s² * t

⇒ v = Vo - a*t = 8.9m/s - 2.744m/s²*t

This gives us:

23.45rad/s² * t = (8.9m/s - 2.744m/s²*t) / 0.11m

2.744*t = 8.9 - 2.744*t

t = 8.9 / 5.488 = 1.622 s ≈ 1.6 s

4) How far does the bowling ball slide before it begins to roll without slipping?

x = Vo*t - ½at² = (8.9*1.622 - ½*2.744*(1.622)²) m = 10.82 m ≈ 11 m

5) What is the magnitude of the final velocity?

v = Vo - at = 8.9m/s - 2.744m/s² * 1.622s = 4.45 m/s

6) After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball:

trans KE = ½ * 4.1kg * (4.45m/s)² =40.595 J ≈ 41 J

I = (2/5)mr² = (2/5) * 4.1kg * (0.117m)² = 0.0224 kg·m²

ω = v/r = 4.45m/s / 0.117m = 38.03 rad/s, so

rot KE = ½Iω² = ½ * 0.0224kg·m² * (38.03rad/s)² = 16.2 J

16.2 J < 41 J

KErot < KEtran

(For a rolling solid sphere, KErot ≈ 2/5 * KEtran)

6 0
3 years ago
Explain why water boiling in a container stops<br>boiling momentarily when the lid is removed​
Dmitriy789 [7]

Answer:

It's because it tips over the threshold from nucleate boiling, which we can see, to convection boiling, which we can't. ... Even if the steam stayed in the pot, it would still stop boiling when you removed the heat. The steam and water in a liquid/vapour mixture are at the same temperature (100ºC).

Explanation:

8 0
3 years ago
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