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faust18 [17]
3 years ago
14

Pls help me i will name brainliest

Physics
2 answers:
Mandarinka [93]3 years ago
7 0

Answer:

a for the first one b for the second

jolli1 [7]3 years ago
6 0
Agreed when looking at the law it will end up with a for firs answer and b for the second
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Which object has the most momentum?
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C) has the most mass
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What happens to air as it warms ,expands and become less dense
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Air expands as it warms. Therefore warm air is less dense than cool air. The warm air from the first floor apartments rises to the second floor. People on the second floor require less heating to keep their <span>apartments comfortable.
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3 years ago
A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?
slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

  • m= 0.500 kg
  • v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

<u><em>Ec= 156.25 J</em></u>

Finally, the bullet has a kinetic energy of 156.25 J.

Learn more about kinetic energy:

brainly.com/question/25959744

brainly.com/question/14028892

#SPJ1

5 0
2 years ago
PLEASE HELP ME A lens with a surface that curves outward like the exterior of a sphere is __________. (Points : 1) reflected ref
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Convex.

Concave curved inward (like how a cave foes in) and convex curves outward. Reflected and refracted do not apply to a lens.
3 0
3 years ago
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A 60kg bicyclist (including the bicycle) is pedaling to the
Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight (W=mg): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

3 0
3 years ago
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