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3 years ago

A sinusoidal wave travels with speed 200 m/s. Its wavelength is 4.0 m. What is its frequency?

Physics

1 answer:

liubo4ka [24]3 years ago

3 0

Answer:

its frequency = 50 per second

Explanation:

$wavelength = speed * time periodtime period = 1 / frequencyfrequency = speed / wavelength\\= 200 / 4.0 = 50 / s$

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In a certain state, electricity is generated by using Earth's heat. Wells are drilled and natural steam is taken out through pip

MArishka [77]

Answer:

Geothermal energy is used. Dry system plants are set up for harnessing this energy.

Explanation:

3 0

3 years ago

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

$tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}$

From Rayleigh criteria

$sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm$

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

$d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm$

The focal length of the camera's lens is 300cm

3 0

3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

6 0

3 years ago

4.Upthrust doesnt depend on the following Physical quantity?

NemiM [27]

Answer:

C: Mass of body

Explanation:

Upthrust is defined as the upward force that a liquid will exert on a body floating atop it.

Now, Formula for upthrust is;

Upthrust = density of liquid × volume of the body × acceleration due to gravity.

From the formula, we can see that upthrust depends on density, volume and acceleration due to gravity.

Thus, looking at the options, the one that doesn't apply is mass of body.

8 0

3 years ago

What is the maximum centripetal acceleration experienced by a person standing still on the surface of the Earth? Where must they

skelet666 [1.2K]

Answer:

The person must be located in the Equator Line. The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

Explanation:

Physically speaking, the centripetal acceleration ( $a_{r}$ ), measured in meters per square second, experienced by a person is defined by the following expression:

$a_{r} = \omega^{2}\cdot r$ (1)

Where:

$\omega$ - Angular speed of the Earth, measured in radians per second.

$r$ - Distance perpendicular to the rotation axis, measured in meters.

Since rotation axis passes through poles and distance described above is directly proportional to centripetal acceleration. The person must be located in the Equator Line, which is equivalent to the radius of the planet.

In addition, the angular speed of the Earth can be calculated in terms of its period ( $T$ ), measured in seconds:

$\omega = \frac{2\pi}{T}$ (2)

If we know that $r = 6.371\times 10^{6}\,m$ and $T = 86400\,s$ , then the maximum centripetal acceleration experienced by a person is:

$a_{r} = \left(\frac{2\pi}{86400\,s} \right)^{2}\cdot (6.371\times 10^{6}\,m)$

$a_{r} = 0.0337\,\frac{m}{s^{2}}$

The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

6 0

3 years ago