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lesantik [10]
3 years ago
14

STATE THE LAWS OF CONSERVATION OF MOMENTUM​

Physics
1 answer:
dalvyx [7]3 years ago
6 0

\Huge  \mid   \underline {\mathcal {{{\color{purple}{Answer...}}}}} \mid

When two bodies collide with each other in the absence of an external force, then the total final momentum of the bodies is equal to their total initial momentum.

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Select the correct answer.
r-ruslan [8.4K]

Answer:

That would be B. Hope this helps!

Explanation:

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3 years ago
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Identify three specific situations in which machines make work easier.
Komok [63]
Construction, like building a home/building, digging, like in a mine, and opening a soda can, where the part to open is a lever.
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3 years ago
1) Momenum is
adelina 88 [10]

Answer:

momentum formula = Mass × Velocity

3 0
3 years ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
3 years ago
Vector question,university physics zemanski
just olya [345]

Answer:

yes This is correct Answer

3 0
2 years ago
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