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viva [34]
3 years ago
7

juie ran eight laps around her school's quarter mile track in 18 minutes. her average velocity was _______m/s

Physics
2 answers:
Blizzard [7]3 years ago
5 0

Answer:

About 3m/s

Explanation:

If Julie ran eight quarter mile laps, then she ran a total of 1/4 * 8=2 miles. 2 miles is about 3220 meters, and 18 minutes is 1080 seconds, meaning that her average velocity was about 3m/s. Hope this helps!

iren2701 [21]3 years ago
3 0
Around 3m/s
That’s what I think but am not sure.
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A .5 kg toy train car moving forward at 3 m/s collides with and sticks to a .8 kg toy car that is traveling at 2 m/s what is the
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Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1}+ m_{2} * v'_{2}

In case of perfectly inelastic collision v'1 and v'2 are same.

We are given information:
m₁=0.5kg
m₂=0.8kg
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2 years ago
A material that has high resistance to the flow of electric current is called an electric ______
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A material that has high resistance to the flow of electric current is called an electric resistor
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Read 2 more answers
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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