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2 years ago

9

If the forces are equal and opposite why does the rocket ship move

1 answer:

qaws [65]2 years ago

7 0

<span>A rocket in its simplest form is a chamber enclosing a gas under pressure. A small opening at one end of the chamber allows the gas to escape, and in doing so provides a thrust that propels the rocket in the opposite direction. A good example of this is a balloon. Air inside a balloon is compressed by the balloon's rubber walls. The air pushes back so that the inward and outward pressing forces are balanced. When the nozzle is released, air escapes through it and the balloon is propelled in the opposite direction.</span>

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2 years ago

According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s

Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

$\tau_{net} = 0$

now we have

$\frac{dL}{dt}= 0$

now we also know that

$Area = \frac{1}{2}r^2d\theta$

so rate of change in area is given as

$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}$

so we will have

$\frac{dA}{dt} = \frac{1}{2}r^2\omega$

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3 years ago

Quick does the watermelon have more or less mass then the 2kg bottle?

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Answer

it will be less\

Explanation:

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3 years ago

When the current through a circular loop is 6.0 A, the magnetic field at its center is 2.0 * 10-4 T. What is the radius of the l

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Answer:

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2 years ago

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto

Fed [463]

Answer:

a) a = 34.375 m / s², b) v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

v_f = $\frac{x-x_1}{t}$

we substitute the values

v_f = $\frac{ 6600 -x_1}{4}$

The initial part of the movement is carried out with acceleration

v_f = v₀ + a t

x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

x₁ = ½ a t²

v_f = a t

we substitute the values

x₁ = 1/2 a 16²

x₁ = 128 a

v_f = 16 a

let's write our system of equations

v_f = $\frac{6600 - x_1}{4}$

x₁ = 128 a

v_f = 16 a

we substitute in the first equation

16 a = $\frac{6600 -128 a}{4}$

16 4 a = 6600 - 128 a

a (64 + 128) = 6600

a = 6600/192

a = 34.375 m / s²

b) let's find the time to reach this height

x = ½ to t²

t² = 2y / a

t² = 2 5100 / 34.375

t² = 296.72

t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

v_f = 16 a

v_f = 16 34.375

v_f = 550 m / s

8 0

2 years ago