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Shtirlitz [24]
3 years ago
10

I'd like some help in these two questions please, thankyou so much. have a great day! stay safe and stay happy. (there are two p

arts of one question so uh yea.

Physics
1 answer:
Andrews [41]3 years ago
6 0

Answer:

The resultant force is 25 N in the negative x-direction.

Explanation:

We need to find the total force in the x and y-direction and then the resultant force will be the magnitude of the sum of Fx and Fy.

<u>Total force in the x-direction:</u>

F_{x}=12-37=-25\: N

<u>Total force in the y-direction:</u>

F_{y}=10-10=0\: N

Therefore, the magnitude of the resultant force will be:

F_{R}=\sqrt{F_{x}^{2}+F_{y}^{2}}

F_{R}=\sqrt{(-25)^{2}+0}

|F_{R}|=25\: N

The resultant force is 25 N in the negative x-direction, in other words, the car will move to the left.

I hope it helps you!

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Explain how energy is conserved when nuclear fission or fusion occurs
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The mass lost in the nuclear reaction is all converted to energy.
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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)

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3 years ago
What power rating of resistors would you use in the application required it to handle<br> 0.6W?
aliya0001 [1]

I would use a resistor rated for 1 W or more. Not less.

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A circuit has a voltage drop of 24.0 V across a 30.0 resistor that carries a
beks73 [17]

(A) 19.2 W

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3 years ago
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For a vehicle to negotiate a banked curve in poor weather conditions, where the force of friction f = 0, for a given velocity v
djverab [1.8K]

Answer:

R=240m

Explanation:

From the question we are told that:

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