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3 years ago

Calculate the hang time of an athlete who jumps a vertical distance of 0.75 meter

1 answer:

7 0

The time it takes going up will be equal the time returning: 
s = s0 + v0t + (1/2)at^2
Taking the top of the jump as 0 and positive direction downward,
h = 0 + 0 + (1/2)gt^2
t = √(2h/g)
Since the hang time will be twice the time to rise or fall, therefore:

T = 2t
T = 2√(2*0.75/9.80665)
T ≈ 0.78 seconds

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Consider the resistances of short and open circuits. Fill in the blanks: The voltage across a short circuit will always be _____

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Answer:

a) Zero

b) power source

Explanation:

According to Ohm's law, the voltage dropped in a resistance is proportional to the current flow and the resistor opposing to it.

$V=I*R$

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3 years ago

A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

Stella [2.4K]

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

$\terxt{time for compression}=\text{time for half revolution}$

$\terxt{time for compression}=\dfrac{1}{2}\times T$

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}$

Put the value into the formula

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60$

$\terxt{time for compression}=0.2\ sec$

We need to calculate the rate of internal energy

Using first law of thermodynamics

$U=Q-W$

$\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}$

Put the value into the formula

$\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)$

$\dfrac{\Delta U}{\Delta t}=6.8\ kW$

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

$\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}$

$\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}$

Put the value into the formula

$\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}$

$\Delta\theta=32.48^{\circ}$

Hence, The temperature change per compression stroke is 32.48°.

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4 years ago

Please help me on question 3a and 3b. Thanks!

Sonbull [250]

(a) The frequency of water wave is 2 Hz.

(b) The wave speed of the water wave is 3.6 m/s.

Explanation:

(a) It is known that completion of one complete wave in 1 second is defined as frequency of 1 HZ. So here there are 120 waves crossing the boat in 1 minute. So the frequency of the water wave will be

$Frequency =\frac{\text { Number of waves }}{\text { Time in seconds }}$

As the time is 1 minute which is equal to 60 seconds and the number of waves is given as 120 then the frequency of the water wave is

$\text { Frequency }=\frac{120}{60}=2 \mathrm{Hz}$

So the frequency of water wave is 2 Hz.

(b) Then if the wavelength of the water wave is 1.8 m with a frequency of 2 Hz, then speed of the wave can be determined as the product of wavelength with frequency.

So Speed = Frequency × Wavelength

Speed = 2 × 1.8 = 3.6 m/s.

So the speed of the water wave is 3.6 m/s.

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3 years ago