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STALIN [3.7K]
3 years ago
13

The speed of the cart after 8 seconds of Low fan speed is

Physics
1 answer:
Gelneren [198K]3 years ago
3 0

Answer:

8 seconds on low = 72.0

3 seconds on medium = 36.0

6 seconds on high = 96.0

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3.
Inga [223]

Answer:

Iris

Explanation:

The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.

7 0
3 years ago
A wave has a frequency of 15,500 Hz and a wavelength of 0.20 m. What is the
Hoochie [10]

Answer:

3100 m/s

Explanation:

The relationship between frequency and wavelength of a wave is given by the wave equation:

v=f\lambda

where

v is the speed of the wave

f is its frequency

\lambda is the wavelength

For the wave in this problem,

f = 15,500 Hz

\lambda=0.20 m

Therefore, the wave speed is

v=(15500)(0.20)=3100 m/s

4 0
3 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed v of the geosynchronous satellite is given by:

v=\frac{2\pi R}{T}

Because 2\pi R is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

F_c=\frac{mv^{2} }{R}

If we substitute the expression for v in this formula, we get:

F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}

Since the centripetal force is the gravitational force F_g between the satellite and the Earth, we know that:

F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }

Where G is the gravitational constant (G=6.67*10^{-11} Nm^{2}/kg^{2}) and M is the mass of the Earth (M=5.97*10^{24}kg). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.

5 0
3 years ago
An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pi
Harman [31]

Answer:

The  length is  l  =  8.6 \  m

Explanation:

From the question we are told that

   The frequencies of the two successive harmonics are f_1 =  220 \ Hz ,  f_2  =  240 \  Hz

   The speed of sound in the air is  v_s  =  343 \  m/s

Generally the frequency of a given harmonic is mathematically represented as

     f_n  =  \frac{n  v  }{2l}

Here  n defines  the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

    220  =  \frac{n v}{2l}

and  

     240  =  \frac{(n+1) v}{2l}

So

   \frac{(n + 1 ) v}{2l} - \frac{n v}{2l}  =  240-220

=>  \frac{v}{2l}  =  20

=>   l  =  8.6 \  m

7 0
3 years ago
If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice
Savatey [412]

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

- Sun (closest to the Earth)

- Betelgeuse.

- Antares.

- Vega.

8 0
3 years ago
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