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gladu [14]
3 years ago
12

What is the power of a light bulb that has a resistance of 190 ohms in a 120 v circuit?

Physics
2 answers:
oksian1 [2.3K]3 years ago
8 0
The answer after you round is
75.8
Alenkinab [10]3 years ago
5 0
The power is 75.78947 your welcome
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A motorcycle, which has an initial linear speed of 8.0 m/s, decelerates to a speed of 2.2 m/s in 4.1 s. Each wheel has a radius
SVETLANKA909090 [29]

Answer:

Explanation:

Given

Initial linear speed v_1=8 m/s

initial angular velocity \omega _1=\frac{v_1}{r}=\frac{8}{0.6}=13.33 rad/s

Speed after 4.1 s is  v_2=2.2 m/s

\omega _2=\frac{2.2}{0.6}=3.66 rad/s

using \omega _2=\omega _1+\alpha t

where \alphais angular acceleration

3.66=13.33+\alpha \cdot 4.1

\alpha =-2.37 rad/s^2 i.e. clockwise

(b)angular displacement

\theta =\omega _1t+\frac{\alpha t^2}{2}

\theta =13.33\times 4.1-\frac{2.37\cdot 4.1^2}{2}

\theta =54.66-19.75

\theta =34.91 rad

3 0
3 years ago
PLEASE HELP
vagabundo [1.1K]
The magnetic field is holding onto the nail that’s all I got
7 0
3 years ago
How many paddles should be on a water wheel?
lesantik [10]
12.... it helps to markout the location of ecah paddle on the wheel
4 0
3 years ago
A ball is projected into the air. Where is potential energy the greatest? A) A. B) B. C) C. D) A. and C.
Ksju [112]

The point with the greatest potential energy is B.

The potential energy of an object is depends on the relative distance between the object and the ground and its mass. The higher the object is from the ground the greater the potential energy posses.

Potential energy (P.E) = mgh

Where m is the mass, g is the gravity and h is the height from the ground to where the object is.

Since the mass and gravity is constant in this case, only the height will determine the point with the greatest P.E and that point is B.

5 0
3 years ago
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You have a mass of 55kg and are riding your frictionless skateboard, which has a mass of 5kg, in a straight line at a speed of 4
Rzqust [24]

Given

m1(mass of the first object): 55 Kg

m2 (mass of the second object): 55 Kg

v1 (velocity of the first object): 4.5 m/s

v2 (velocity of the second object): ?

m3(mass of the object dropped): 2.5 Kg

The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:

Pa= Pb

Where Pa is the momentum before collision and Pb is the momentum after collision.

Now applying this law for the above problem we get

Momentum before collision= momentum after collision.

Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s

Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2

Now we know that Momentum before collision= momentum after collision.

Hence we get

270 = 62.5 v2

v2 = 4.32 m/s




7 0
3 years ago
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