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pychu [463]
3 years ago
11

A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i

s horizontal and 8.3 m above the other shore wherethe car lands. The tires on the car all hit at once and the air resistance is insignificant.How long is the car in the air?
Physics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:1.301 s

Explanation:

Given

Initial Velocity(u)=30 m/s

Height of cliff=8.3 m

Time taken to cover 8.3 m

h=ut+\frac{at^2}{2}

here Initial vertical velocity is 0

8.3=\frac{gt^2}{2}

t^2=1.69

t=1.301 s

Horizontal distance

R=u\times t

R=30\times 1.301=39.04 m

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A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.
Galina-37 [17]

Answer:

Change in specific internal Energy=250\ \rm Btu/lb

Explanation:

Given:

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  • Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\

So the final volume of the gas is calculated.

Work in any isotropic is given by w

w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\

According to the first law of thermodynamics we have

Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\

So the Specific Internal Change is given by

\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb

So the specific Change in Internal energy is calculated.

6 0
3 years ago
Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black
oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

Q = FA\sigma\Delta T^4

Where,

F =View Factor

A = Cross sectional Area

\sigma = Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K

The view factor between two coaxial parallel disks would be

\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33

\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75

Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

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5 0
3 years ago
Examine the circuit. Pretend you are an electron flowing through this circuit and you are with a group of other electrons. Sudde
melamori03 [73]

You knew that this question is ridiculously easy.  So, just to
make it harder, you decided not to let us see the picture, so
that we could not "examine the circuit".

The description is talking about a parallel circuit.  The other
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We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

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a'=4\times a

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