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disa [49]
2 years ago
5

A 2.00-kg block of ice is moving on a frictionless horizontal surface. At t = 0 the block is moving to the right with a velocity

of magnitude 3.00 m/s. (a) Calculate the velocity of the block (magnitude and direction) after a force of 5.00 N directed to the right has been applied for 4.00 s; (b) if instead a force of 7.00 N directed to the left is applied from t=0 to t = 4.00 s, what is the final velocity of the block?
Physics
1 answer:
enyata [817]2 years ago
4 0

Answer:

Explanation:

a )

We shall apply the concept of impulse .

Impulse = force x time = change in momentum

= 5 x 4 = 2 ( V - 3 )  , where V is final velocity of the object

20 = 2V - 6

V = 13 m /s

b )

Impulse applied = - 7 x 4 = - 28 kg m/s ( negative as direction of force is opposite motion )

If v be the final velocity

2 x 3 - 28 = 2 v  ( initial momentum - change in momentum = final momentum )

- 22 = 2v

v = - 11 m /s

object will move with 11 m /s in opposite direction .

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Answer:

Astronomers have no theoretical explanation for the ""hot Jupiters"" observed orbiting some other stars.

False

Explanation:

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2 years ago
3.2 Define conservation of energy.<br>​
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A principle stating that energy cannot be created or destroyed, but can be altered from one form to another.

Explanation:

5 0
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Which portion of the electromagnetic spectrum can the human eye detect? *
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Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?
MrMuchimi

D. Redefine the machine’s system boundaries.

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7 0
3 years ago
Read 2 more answers
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
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