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Rasek [7]
3 years ago
10

What is half the vertical distance from the crest to the trough of a transvers wave?​

Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

It is the amplitude.

Explanation:

The amplitude is the measure of the size of the disturbance from a wave, so half the distance from the trough and crest.

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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°.

6 0
4 years ago
An underwater diver sees the sun 50° above horizontal. how high is the sun above the horizon to a fisherman in a boat above the
Usimov [2.4K]
Snell's law states that:

n1 Sin∅1 = n2 Sin ∅2

Where, medium 1 with (n1 = 1.33) is water and medium 2 with (n2 = 1) is the air, ∅1 = 90-50 = 40°

Therefore,
Sin ∅2 = n1/n2 *Sin ∅1 = 1.33/1 *Sin 40 = 0.4833=> ∅1 = Sin ^- (0.4833) = 28.9 °

The fisherman the sun at 61.1° (90-∅2) above the horizontal.
3 0
3 years ago
A 2 kg ball of clay moving at 35 m/s strikes a 10 kg box initially at rest. What is the velocity of the box after the collision?
Ainat [17]

Answer:

V = 5.83 m/s

Explanation:

Given that,

Mass of a ball of a clay, m = 2 kg

Initial speed of the clay, u = 35 m/s

Mass of a box, m' = 10 kg

Initially, the box was at rest, u' = 0

We need to find the velocity of the box after the collision. Let V be the common speed. Using the conservation of momentum to find it.

mu+m'u'=(m+m')V\\\\V=\dfrac{mu+m'u'}{(m+m')}\\\\V=\dfrac{2\times 35+10\times 0}{2+10}\\\\V=5.83\ m/s

So, the velocity of the box after the collision is equal to 5.83 m/s.

4 0
3 years ago
A .5kg bird is perched on its nest so that it has 50J of potential energy. how far is it off the of the ground?
pshichka [43]

It is 10.20 m from the ground.

<u>Explanation:</u>

<u>Given:</u>

m = 0.5 kg

PE = 50 J

We know that the Potential energy is calculated by the formula:

P. E = m \times g \times h

where m is the is mass in kg;  g  is acceleration due to gravity which is 9.8 m/s  and  h  is height in meters.

PE is the Potential Energy.

Potential Energy is the amount of energy stored when an object is stationary.

Here, if we substitute the values in the formula, we get

P. E = m \times g \times h

50 = 0.5 × 9.8 × h

50 = 4.9 × h

h = \frac {50} {4.9}

h = 10.20 m

3 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
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