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soldi70 [24.7K]
3 years ago
5

Which statement best compares the momentum of a 5-kilogram fish swimming at a speed of 10 meters/second and a 2-kilogram fish sw

imming at a speed of 10 meters/second?
Physics
2 answers:
ruslelena [56]3 years ago
6 0
M = mass of the larger fish =5kg
<span>V = velocity of the larger fish =10m/s</span>
<span>m = mass of the smaller fish =2kg</span>
<span>v = velocity of the smaller fish =10m/s
</span>formula=
<span>MV = mv 
5kg*10m/s=2kg*10m/s
biggern mass fish has more momentum
hope this helps

</span>
Ronch [10]3 years ago
5 0

Answer:

The momentum of the 5-kilogram fish is greater than the momentum of the 2-kilogram fish

Explanation:

The momentum of each fish can be calculated as follows:

p=mv

where

m is the mass of the fish

v is the velocity of the fish

- For the first fish, we have:

p=mv=(5 kg)(10 m/s)=50 kg m/s

- For the second fish:

p=mv=(2 kg)(10 m/s)=20 kg m/s

Therefore, the momentum of the first fish is greater than the second fish.

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Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
Drupady [299]

Answer:

14.57 ohms

Explanation:

Here in the figure ,Rb & R₄are in series  & also  Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )

                                          = 20 ║ 15

                                          = (20×15) / (20 + 15)

                                          = 8.57 ohms

Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.

5 0
3 years ago
Look at the graph pic and answer the question correctly!
SCORPION-xisa [38]

Answer:

B)

Explanation:

That the time period of which they stop.

4 0
3 years ago
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Students have been assigned to write reports on cell organelles. Eric’s report is about the organelle that supports and gives st
mojhsa [17]
Eric is writing about the cell wall.
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The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an
9966 [12]

Answer:

-4.941*10^8J.

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

M_e = 5.97*10^{24}Kg

r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

4 0
3 years ago
Ur a genius if u explain how it’s A correctly
Alinara [238K]

Answer:

Explanation:

4/1=4

3/2=1.5

2/3=0.666667

1/4=0.25

D has the least number so its D

6 0
3 years ago
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