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3 years ago

A 7.0kg skydiver is descending with a constant velocity

Physics

1 answer:

Pavel [41]3 years ago

4 0

Answer:

Air resistance: 68.6 N

Explanation:

For the skydiver falling down, there are two forces acting on him:

- The force of gravity, of magnitude

$W=mg$

where

m = 7.0 kg is the mass of the skydiver

$g=9.8 m/s^2$ is the acceleration of gravity

This force acts in the downward direction

- The air resistance, R, in the upward direction

So the net force on the skydiver is:

$\sum F=mg-R$

According to Newton's second law of motion, the net force is also equal to mass times acceleration:

$\sum F=ma$

However, in this problem the skydiver is falling at constant velocity, so his acceleration is zero:

$a=0$

Therefore, the net force is zero:

$\sum F=0$

And so we can find the magnitude of the air resistance, which is equal to the weigth of the skydiver:

$mg-R=0\\R=mg=(7.0)(9.8)=68.6 N$

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Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

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The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
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In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
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And now we can use the previous equation to calculate the field strength at that altitude:
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3 years ago

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3 years ago

1. A 4000-kg truck traveling with a velocity of 20 m/s due south collides headon with a 1350-kg car traveling with a velocity of

velikii [3]

(a) The momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) The size of momentum is 93500 kgm/s and it will be directed towards South.

Explanation:

The mass of the truck moving due south is given as 4000 kg and the speed is 20 m/s. Similarly, the mass of the car moving due north is 1350 kg and the speed is 10 m/s.

(a) Then the momentum of each vehicle can be obtained by the product of mass with their respective speed.

$Momentum of truck = Mass * Speed = 4000 * 20 =80000 kgm/s$

Similarly, the momentum of car will be

$Momentum of car = 1350 * 10 = 13500 kgm/s$

So, the momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) Since, after collision, the vehicles stick together, the momentum after collision will be equal to the total momentum of both the vehicles before collision. This is because, it will obey conservation of momentum.

Momentum of vehicles after collision = total momentum before collision

Momentum after collision = 80000+13500 = 93500 kgm/s.

The direction of the vehicles after collision will be towards south as the mass and speed of the truck is greater than car. So the impact or force exerted on the car by the truck will be greater and thus both the vehicles will be directed towards south after collision.

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3 years ago

A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

$dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5$

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

$W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J$

W = 3.12 J

Hope this helps!

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