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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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Answer: i d k

Explanation:

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2 years ago
A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio
Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

F - External force exerted on the sled, measured in newtons.

f - Friction force, measured in newtons.

N - Normal force from the ground on the mass, measured in newtons.

W - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

If we know that m = 50\,kg and g = 9.807\,\frac{m}{s^{2}}, the weight of the sled is:

W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

W = 490.35\,N

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

F_{min,s} = \mu_{s}\cdot W (2)

Where:

\mu_{s} - Static coefficient of friction, dimensionless.

W - Weight of the sled, measured in newtons.

If we know that \mu_{s} = 0.3 and W = 490.35\,N, then the force needed to start the sled moving is:

F_{min,s} = 0.3\cdot (490.35\,N)

F_{min,s} = 147.105\,N

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

F_{min,k} = \mu_{k}\cdot W (3)

Where \mu_{k} is the kinetic coefficient of friction, dimensionless.

If we know that \mu_{k} = 0.1 and W = 490.35\,N, then the force needed to keep the sled moving at a constant velocity is:

F_{min,k} = 0.1\cdot (490.35\,N)

F_{min,k} = 49.035\,N

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0
2 years ago
Astronomers estimate that a 2
tia_tia [17]
M)³ / 6 = 4.2e9 m³ 
<span>so its mass is </span>
<span>M = 3300kg/m³ * 4.2e9m³ = 1.4e13 kg </span>
<span>and so its KE at 16 km/s = 16000 m/s is </span>
<span>KE = ½ * 1.4e13kg * (16000m/s)² = 1.8e21 J 

</span># of bombs N = 1.8e21J / 4.0e16J/bomb = 44 234 bombs 
<span>give or take. 
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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

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3 years ago
What is one way to induce an electric current
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<span>1.an electric is induced when you move a magnet through a coil wire

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Why does increasing the number of trials increase confidence in the results of the experiment?
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It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.
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