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Nonamiya [84]
2 years ago
11

Alex, parked by the side of an east-west road, is watching car P, which is moving in a westerly direction. Barbara, driving east

at a speed 52 km/h, watches the same car. Take the easterly direction as positive. If Alex measures a speed of 78 km/h for car P, what velocity will Barbara measure?
Physics
1 answer:
Alexxandr [17]2 years ago
6 0

Answer:

v_{PB} = 130\ km/h

Explanation:

Since, Alex is at rest. Therefore, the speed measured by him will be the absolute speed of car P. Therefore, taking easterly direction as positive:

Absolute\ Velocity\ of\ Car\ P = v_{P} = -78\ km/h

And the absolute velocity of Barbara's Car is given as:Absolute\ Velocity\ of\ Barbara's\ Car = v_{B} = 52\ km/h

Now, for the velocity of Car p with respect to the velocity of Barbara's Car can be given s follows:

Velocity\ of\ Car\ P\ measured\ by\ Barbara = v_{PB} = v_{B}-v_{P}\\\\v_{PB} = 52\ km/h-(-78\ km/h)

v_{PB} = 130\ km/h

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The Earth and Moon are separated by about 4.0 10^6 m. Suppose Mars is [{MathJax fullWidth='false' 2.9 \times 10^{11} }] m from E
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a) α=7.9x10^-4 rad

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Explanation:

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4 0
3 years ago
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!!!

3 0
3 years ago
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const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

5 0
3 years ago
Need help solving this
Nostrana [21]

Answer:

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Explanation:

5 0
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