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lbvjy [14]
3 years ago
15

Model rocket engines are rated by the impulse that they deliver when they fire. A particular engine is rated to deliver an impul

se of 3.5 kg #m/s. The engine powers a 120 g rocket, including the mass of the engine. What is the final speed of the rocket once the engine has fired?
Physics
1 answer:
abruzzese [7]3 years ago
4 0

Answer:

Final speed of the rocket, \Delta v= 29.16\ m/s

Explanation:

It is given that,

Impulse delivered by an engine, J = 3.5 kg-m/s

Mass of the rocket, m = 120 g = 0.12 kg

To find,

The final speed of the rocket

Solution,

We know that the impulse is equal to the product of mass and velocity o it is equivalent to the change in momentum of an object.

J=\Delta P=m\times \Delta v

\Delta v=\dfrac{J}{m}

\Delta v=\dfrac{3.5\ kg-m/s}{0.12\ kg}

\Delta v= 29.16\ m/s

So, the final speed of the rocket once the engine has fired is 29.16 m/s.

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Answer:

Particles in a water wave exchange kinetic energy for potential energy. When particles in water become part of a wave, they start to move up or down. This means that kinetic energy (energy of movement) has been transferred to them

Explanation:

hope this helps u ....

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6 0
3 years ago
In which medium would sound travel the fastest? a)across a room b)in a swimming pool c)through outer space d)through a railroad
lora16 [44]

d)through a railroad track

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A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg 
Aleonysh [2.5K]

Heat required to change the phase of ice is given by

Q = m* L

here

m = mass of ice

L = latent heat of fusion

now we have

m = 45 kg

L = 334 KJ/kg

now by using above formula

Q = 45 * 334 * 10^3

Q = 1.5 * 10^7 J

In KJ we can convert this as

Q = 1.5 * 10^4 kJ

so the correct answer is D option

7 0
3 years ago
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A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point
Alexus [3.1K]

Answer:

(2.5,0)

Explanation:

The particle can be described by the following equations:

x=Rsin(-\omega t)+2.5\\y=Rcos(-\omega t)\\\frac{dx}{dt}=-\omega Rcos(-\omega t)\\\frac{dy}{dt}=\omega Rsin(-\omega t)\\\frac{d^2x}{dt^2}=-\omega^2Rsin(-\omega t)\\\frac{d^2y}{dt^2}=-\omega^2Rcos(-\omega t)

For R = 2.5, ω = 2 and t = 0:

x=2.5\\y=2.5\\\\\frac{dx}{dt}=-5\\ \frac{d^2y}{dt^2}=-10

The center of the circle would be at point (2.5,0)

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3 years ago
In a rear-end collision, occupants are thrown towards the __________ of the vehicle.
Harrizon [31]

In a rear-end collision, occupants are thrown towards the BACK of the vehicle.

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