A basketball player throws a ball horizontally toward another player 10 meters away. If the ball is

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

N

9966 [12]

Answer: de?

Explanation:

7 0

3 years ago

A constant tone is being applied to a speaker. The voltage across the speaker is 5 volts. The voltage across the speaker is incr

lubasha [3.4K]

Answer:

decibel gain is 10 db

Explanation:

given data

voltage v1 = 5 volts

voltage v2 = 15 volts

to find out

decibel gain

solution

we have given that 5 volt voltage that is increase to 15 volts

so

we know here decibel gain formula that is

decibel gain = 20 log(v2/v1) ....................1

put here value of v1 and v2 in equation 1

decibel gain = 20 log(15/5)

decibel gain = 20 log3

decibel gain = 20 × 0.4771

decibel gain = 9.54 db = approx 10 db

so decibel gain is 10 db

5 0

4 years ago

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What Type of relationship exists between the Temperature of a Star and the Wavelength of a Star?

Debora [2.8K]

This is known as Wien's Law:
The relationship is:
wavelength = 0.0029/temperature

It is an inversely proportional relationship.

3 0

4 years ago

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Which statements accurately describe the shape of Earth’s orbit around the Sun? Check all that apply.

damaskus [11]

<span>The Earth’s orbit is a nearly circular ellipse.
</span><span>The Sun is located at one of the two focal points.</span>

The Earth moves around the Sun in an orbit that is almost but not quite circular. As Kepler proved in the seventeenth century, the orbit is actually an ellipse. A parameter called the eccentricity (e) defines the degree of departure from a circle. A value of e=0 would indicate a circle whereas a value of e=0.9 would indicate a very elongated ellipse. The eccentricity of the Earth's orbit is currently e=0.0167.

4 0

3 years ago

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