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3 years ago

Which situation illustrates Newton's first law?

1 answer:

7 0

Newton's first law states that - An object in motion will stay in motion unless compelled to change it's state by the action of an external force.

Thus, your answer is D.

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The height of a circular tower is 179 meters and its diameter is 48 meters. (Assume that the building is a perfect cylinder). Wh

EastWind [94]

The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:

$h = \frac{179m}{166} = 1.078m = 107.8cm$

The same for the diameter,

$\phi = \frac{48m}{166}= 0.2891m = 28.91cm$

The volume of a cylinder is given as

$V = (\pi r^2)(h)$

$V = (\pi (\frac{d}{2})^2)(h)$

$V = (\pi (\frac{ 28.91}{2})^2)(107.8)$

$V = 70762.8cm^3$

Therefore the volume would be $V = 70762.8cm^3$

6 0

3 years ago

Acceleration is rate of change of <br>A-Position<br>B-Time<br>C-Velocity<br>D-Speed

Drupady [299]

<h3>Answer:</h3>

[C] Velocity.

<h3>Explanation:</h3>

<u>As we know that</u>,

a = v - u/t

<u>where, a = acceleration, v = final velocity, u = initial velocity and t = time taken to travel</u>.

8 0

2 years ago

Read 2 more answers

If it takes 100 N to move a box 5 meters, what is the work done on the box?

inysia [295]

Answer:

Explanation:

Work=force×distance

W=F×S

=100 N × 5 m

= 500 Nm

W = 500J

7 0

2 years ago

A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci

Vlad [161]

Answer:

Approximately $\rm 2.5\; m \cdot s^{-1}$ .

Explanation:

Let the increase in the rocket's velocity be $\Delta v$ . Let $v_0$ represent the initial velocity of the rocket. Note that for this question, the exact value of $v_0$ doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

Mass of the rocket with the 5 kg of fuel: $1000$ .
Initial velocity of the rocket and the fuel: $v_0$ .
Hence the initial momentum of the rocket: $1000\,v_0$ .

Mass of the rocket without that 5 kg of fuel: $1000 - 5 = 995$ .
Final velocity of the rocket: $v_0 + \Delta v$ .
Hence the final momentum of the rocket: $995\,(v_0 + \Delta v)$ .

Mass of the 5 kg of fuel: $5$ .
Final velocity of the fuel: $v_0 - 500$ (assuming that the the 500 m/s in the question takes the rocket as its reference.)
Hence the final momentum of the fuel: $5\,(v_0 - 500)$ .