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3 years ago

Which hand is negatively charged? A B C D

Physics

2 answers:

vichka [17]3 years ago

6 0

Answer:

the one that has more negative sights so the first hand on your right :)

VLD [36.1K]3 years ago

4 0

Answer:

Explanation:

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3 years ago

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4 years ago

Read 2 more answers

Ill give brainliest! 1. Which item below describes a quick change to

TiliK225 [7]

Answer:

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Explanation:

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3 years ago

A 3 kg block is released from rest to slide down a ramp with friction. The length that the block slides is 2 meters and the angl

Arte-miy333 [17]

Answer:

(a). The acceleration is 3.20 m/s².

(b). The amount of work done by each force is 19.2 J.

(c). The total work done on the block is 19.2 J.

(d). The final speed of the block is 6.26 m/s.

Explanation:

Given that,

Mass of block = 3 kg

Distance = 2 m

Angle = 30°

Coefficient of kinetic friction = 0.20

(a). We need to calculate the acceleration

Using balance equation of force

$N = mg\co\theta$

$mg\sin\theta-f_{k}=ma$

$a = g\sin\theta-\mu g\cos30$

Put the value into the formula

$a=g(\sin30-0.20\cos30)$

$a=9.8(\dfrac{1}{2}-0.20\times\dfrac{\sqrt{3}}{2})$

$a=3.20\ m/s^2$

The acceleration is 3.20 m/s².

(b). We need to calculate the amount of work done by each force

Using formula of work done

Normal force is

$N = mg\cos30$

So due to this the net force is zero then the no work done by reaction force.

By another force,

$W= F\times d$

$W=ma\times d$

Put the value into the formula

$W= 3\times3.20\times2$

$W=19.2\ J$

The amount of work done by each force is 19.2 J.

(c). We need to calculate the total work done on the block

The total work done on the block is 19.2 J.

(d). We need to calculate the final speed of the block

Using equation of motion

$v^2=u^2+2gs$

Put the value into the formula

$v^2=0+2\times9.8\times2$

$v=\sqrt{2\times9.8\times2}$

$v=6.26\ m/s$

The final speed of the block is 6.26 m/s.

Hence, This is the required solution.

4 0

3 years ago