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Fofino [41]
3 years ago
5

Relating to Newton’s Third Law, action and reaction, what is the reaction when a rocket expels gas, smoke and flames from the no

zzle end of the rocket engine?
Physics
1 answer:
Cloud [144]3 years ago
7 0

Answer:

Propels in the opposite direction

Explanation:

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A 2.3 kg block of copper is heated at atmospheric pressure such that its temperature increases from 6 oC to 90 oC. How much heat
Shtirlitz [24]

Answer:

the  heat absorbed by the block of copper is 74368.476J

Explanation:

Hello!

To solve this problem use the first law of thermodynamics that states that the heat applied to a system is the difference between the initial and final energy considering that the mass and the specific heat do not change so we can infer the following equation

Q=mCp(T2-T1)

Where

Q=heat

m=mass=2.3kg

Cp=0.092 kcal/(kg C)=384.93J/kgK

T2=Final temperatura= 90C

T1= initial temperature=6 C

solving

Q=(2.3kg)(384.93\frac{J}{kgC} )(90C-6C)=74368.476J

the  heat absorbed by the block of copper is 74368.476J

7 0
3 years ago
An engine has an energy input of 125 J, and 35 J of that energy is transformed into useful energy. What is the efficiency of the
Korolek [52]
Efficiency = (useful output) / (input)

Efficiency = (35 J) / (125 J) = 0.28  =  28%
8 0
3 years ago
Read 2 more answers
What would happen if your stomach wasint working properly
adelina 88 [10]
Your stomach, as in JUST your stomach?
Well the role of your stomach is to break down large clumps of food. Without that it would be very hard to impossible to digest food.<span />
7 0
3 years ago
Please select the word from the list that best fits the definition
almond37 [142]
The broadest level of organization would be the domain
4 0
3 years ago
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A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-
joja [24]

Answer:

5.25\cdot 10^{40} kg m^2/s

Explanation:

The angular momentum of the pulsar is given by:

L=m\omega r^2

where

m=2.8\cdot 10^{30} kg is the mass of the pulsar

r = 10.0 km = 1\cdot 10^4 m is the radius

\omega is the angular speed

Given the period of the pulsar, T=33.5\cdot 10^{-3} s, the angular speed is given by

\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s

And so, the angular momentum is

L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s

8 0
3 years ago
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