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3 years ago

7

A rocket burns fuel at a rate of 199 kg/s and exhausts the gas at a relative speed of 9 km/s. Find the thrust of the rocket. Ans

wer in units of MN.

1 answer:

andreyandreev [35.5K]3 years ago

7 0

Answer:

1.791 MN

Explanation:

Thrust of the rocket can be found using the relation

T = v.dm/dt, where

T = thrust off the rocket

v = speed of the rocket, 9 km/s = 9000 m/s

dm/dt = rate at which fuel burns, 199 kg/s

Substituting the values into the formula, we have

T = 9000 * 199

T = 1791000 N

T = 1.791*10^6 N

Since 1 MN = 10^6, thus

T = 1.791 MN

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If the car ha a mass of 1000 kilograms, what is its momentum (v=35m/s)

Black_prince [1.1K]

Answer:

<em>The momentum of the car is 35,000 kg.m/s</em>

Explanation:

<u>Momentum</u>

Momentum is often defined as <em>mass in motion.</em>

Since all objects have mass, if it's moving, then it has momentum. It can be calculated as the product of the mass by the velocity of the object:

$\vec p = m\vec v$

If only magnitudes are considered:

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2 years ago

The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulu

Gemiola [76]

Answer:

$\displaystyle w=3.478\ rad/sec$

$M=0.0182\ J$

$v=0.398\ m/s$

Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

$\displaystyle w=\sqrt{\frac{g}{L}}$

We have the length of the pendulum is L=0.81 meters, then we have

$\displaystyle w=\sqrt{\frac{9.8}{0.81}}$

$\displaystyle w=3.478\ rad/sec$

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

$U=mgh$

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

$H+Y=L$

And

$H=L\ cos\alpha$

Solving for Y

$Y=L(1-cos\alpha )$

$Since\ \alpha=8.1^o, L=0.81\ m$

$Y=0.0081\ m$

The potential energy is

$U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)$

$U=0.0182\ J$

The mechanical energy is, then

$M=K+U=0+U=U$

$M=0.0182\ J$

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

$\displaystyle K=\frac{mv^2}{2}$

Equating to the mechanical energy of the system (M)

$\displaystyle \frac{mv^2}{2}=0.0182$

Solving for v

$\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}$

$v=0.398\ m/s$

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3 years ago

You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, $\omega = 25.13 rad/s$

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = $\frac{1}{T}$

f = $\frac{1}{0.25}$

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

$\omega = 2\pi \times f$

$\omega = 2\pi \times 4$

$\omega = 25.13 rad/s$

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3 years ago