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3 years ago

Circuit A

Physics

2 answers:

Licemer1 [7]3 years ago

3 0

By V=IR
A: 24=I*20
I = 1.2A

B: 220 = I*250
I = 0.88A

C: 6= I*3
I = 2 A

C,A,B

I am Lyosha [343]3 years ago

3 0

Answer:

C,A,B

Explanation:

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An air track car with a mass of 6 kg and velocity of 4 m/s to the right collides with a 3 kg car moving to the left with a veloc

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Remember that moment before collision is equal to the moment after collision.

$(m1 \times u1) + (m2 \times u2) = (m1 \times v1) + (m2 \times u2)$

Plugging in our values,

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2 years ago

A weight lifter lifts a 345 N set of weights from ground level to a position over his head, a vertical distance of 1.89 m. How m

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Explanation:

Given

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3 years ago

Which of the following best represents potential energy being converted to kinetic energy? A. A man jogs and stops to drink an e

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3 years ago

Read 2 more answers

A) What minimum velocity must a roller coaster have such that the riders don’t fall out at the top of a loop with a radius of 12

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Explanation:

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2 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

3 years ago