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3 years ago

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Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

of 30.0 m. During the collision at the bottom of the elevator shaft, a 86.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision

1 answer:

Elena L [17]3 years ago

8 0

Explanation:

Below is an attachment containing the solution.

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If there were no dark energy in the universe, the value of __________ would solely determine the evolution and fate of the unive

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Ωmass

Explanation:

If there were no dark energy in the universe, the value of Ωmass would determine the evolution and fate of the universe.

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The universe is filled with photons left over from the Big Bang that today have an average energy of about 4.9 ✕10-4 (correspond

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Answer:

The number of available energy is $4.820\times10^{45}$

Explanation:

Given that,

Energy $E=4.9\times10^{-4}\ J$

Temperature = 2.7 K

Energy states per unit volume $dE=4\times10^{-5}\ eV$

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Using formula of energy

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Where, h = Planck constant

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E = energy

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$N=\dfrac{8\pi\times(4.9\times10^{-4})^2\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}$

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Hence, The number of available energy is $4.820\times10^{45}$

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I think this is the answer:

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2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos

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Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh$

there, no vertical movement

So, on neglect of height term

$P_{1}=P+\dfrac{1}{2}\rho v_{1}^2$

Where, P= atmospheric pressure

$\rho$ = air density

v = speed

Put the value in the equation

$P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)$

$P_{1}=2126.0\ lb/ft^2$

$P_{1}=\dfrac{2126.0}{144}$

$P_{1}= 14.76\ psi$

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{2}=P+\dfrac{1}{2}\rho v_{2}^2$

Put the value in the equation

$P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)$

$P_{2}=2246.17\ lb/ft^2$

$P_{2}=\dfrac{2246.17}{144}$

$P_{2}= 15.59\ psi$

(c). Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{3}=P+\dfrac{1}{2}\rho v_{3}^2$

Put the value in the equation

$P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)$

$P_{3}=2257.93\ lb/ft^2$

$P_{3}=\dfrac{2257.93}{144}$

$P_{3}= 15.68\ psi$

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

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