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GrogVix [38]
3 years ago
9

Q: Which direction are the force vectors pointing for a bullet fired from a gun. Assume that the bullet is traveling to the righ

t at 750 m/s. (the bullets vector)
A: Up, Down, Left

B: Right, Up

C: Right, Down, Left

D: Left, Down


Hope someone sees this ♡ please?
Physics
2 answers:
Anit [1.1K]3 years ago
5 0

Answer:

D: Left, Down

Explanation:

Assuming that the bullet has already left the gun, there are only two forces acting on the bullet now, during its motion:

- The force of gravity (the weight of the object), acting downward

- The air resistance acting against the direction of motion (so, if the bullet is moving right, the air resistance acts to the left)

As a result of these two forces:

- The bullet will accelerates downward in the vertical direction, due to the force of gravity

- The bullet will decelerated (slow down) along the horizontal direction, due to the air resistance that acts against the direction of motion of the bullet

So, the bullet will acquire a parabolic path.

Therefore, the correct answer is

D: Left, Down

REY [17]3 years ago
5 0

Answer:

The correct answer is D.

Explanation:

The force vectors for a bullet fired from a pistol, which moves to the right at 750 m/s are directed to the left and down. This happens because the forces acting on it are gravity and air resistance, the force of gravity has a downward direction and the force of air resistance has a leftward direction, contrary to the direction of the bullet.

Have a nice day!

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Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

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