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3 years ago

Write a brief essay describing how Newton’s Laws explain how a rocket in space can move objects

2 answers:

5 0

The First Law describes how an object acts when no force is acting upon it. So, rockets stay still until a force is applied to move them. Likewise, once they're in motion, they won't stop until a force is applied. Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry. Newton's Third Law states that "every action has an equal and opposite reaction". In a rocket, burning fuel creates a push on the front of the rocket pushing it forward.

BaLLatris [955]3 years ago

3 0

Answer:

Explanation:

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The Surface Pressure at Leh, Ladakh is 800 mb. Now, assuming that Leh is at an altitude of 3500 m and every 100 m increase in he

Basile [38]

We have that the sea level pressure for Leh area is 1150mb mathematically given as

Ps= 1150 mb

<h3> Sea level pressure</h3>

Question Parameters:

Ladakh is 800 mb.

<u>assuming </u>that Leh is at an altitude of 3500 m and every 100 m

increase in height with respect to sea level corresponds to 10 mb pressure,

Generally, for 3500m the pressure change will be 350 mb.

Therefore, here for the sea level <em>pressure</em> we need to add,

Ps=800+350

Ps= 1150 mb

For more information on Pressure visit

brainly.com/question/25688500

8 0

2 years ago

Which statement correctly describes these electric field lines

JulijaS [17]

Answer:

Explanation:

the correct answer is C please follow me

8 0

3 years ago

Read 2 more answers

A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?

denis23 [38]

Answer:

Workdone = 465766038 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

7 0

3 years ago

A 150-W lamp is placed into a 120-V ac outlet. What is the peak current?

devlian [24]

Answer:

Explanation:

Formula

W = I * E

Givens

W = 150

E = 120

I = ?

Solution

150 = I * 120 Divide by 120

150/120 = I

5/4 = I

I = 1.25

Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)

On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.

However I would go with the other answerer's post and multiply both values by √2

3 0

4 years ago

Read 2 more answers

We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth

Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v = $\frac{m}{m+M} \ v_o$

for when n balls have collided

v = $\frac{m}{n \ m + M}$ v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

p₀ = m v₀

final moment

p_f = m v_{1f} + M v_{2f}

p₀ = p_f

m v₀ = m v_{1f} + M v_{2f}

m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

K₀ = K_f

½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

v₁ = v_{1f} v₂ = v_{2f}

m (v₀² - v₁²) = M v₂²

let's use the identity

(a² - b²) = (a + b) (a-b)

we write our equations

m (v₀ - v₁) = M v₂ (1)

m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

m (v₀ - v₁) = M (v₀ + v₁)

v₀ (m -M) = (m + M) v₁

v₁ = $\frac{m-M}{m + M}$ v₀

we substitute in equation 1 to find v₂

$\frac{M}{m}$ v₂ = v₀ - $\frac{m-M}{m+M}$ v₀

v₂ = $\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o$

v₂ = $\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o$

v₂ = $\frac{2m}{m +M} \ v_o$

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0

3 years ago