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3 years ago

Anothereeeeee freee points cuz why not

Physics

2 answers:

Dmitry [639]3 years ago

5 0

Answer:

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Explanation:

Aloiza [94]3 years ago

4 0

Answer:

thx! :D

Explanation:

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30pts what type of energy is feeling the air around you?

miv72 [106K]

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2 years ago

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ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the

antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = $M_A=5 kg$

Velocity of the ball A before collision: $U_A=20 m/s$

Velocity of ball A after collision= $V_A=10 m/s$

Mass of ball B= $M_B$

Velocity of the ball B before collision: $U_B=10 m/s$

Velocity of ball B after collision= $V_B=15 m/s$

$M_AV_A+M_BV_B=M_AU_A+M_BU_B$

$5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s$

$M_B=10kg$

The mass of ball B is 10 kg.

8 0

2 years ago

Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we

sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

Given the following data:

Mass 1 = 80 kg

Velocity 1 = 2 m/s east.

Mass 2 = 70 kg

Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

$Momentum = mass \times velocity$

For Football player 1:

$Momentum = 80 \times 2$

Momentum 1 = 160 Kgm/s.

For Football player 2:

$Momentum = 70 \times 3$

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

$Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160$

Total momentum = 50 Kgm/s.

Note: We subtracted because the football players were moving in opposite directions.

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2 years ago

A physical count of merchandise inventory on november 30

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It depends on the indirect taxes and the share market values

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3 years ago

Potassium ions (K+) move across a 7.0 -mm- thick cell membrane from the inside to the outside. The potential inside the cell is

Reil [10]

Explanation:

Relation between potential energy and charge is as follows.

U = qV

or, $\Delta U = q \times \Delta V$

= $1.6 \times 10^{-19} \times 70 \times 10^{-3}$

= $112 \times 10^{-22}$ J

or, = $1.12 \times 10^{20} J$

Therefore, we can conclude that change in the electrical potential energy $\Delta U$ is $1.12 \times 10^{20} J$ .

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3 years ago