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Margaret [11]
3 years ago
14

Anyone know this... pls help its for a grade​

Physics
1 answer:
hodyreva [135]3 years ago
3 0
His average speed is 45 miles an hour
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Which statement applies only to electric force instead of both electric and magnetic
garri49 [273]

Answer:

C

Explanation:

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Minchanka [31]

Answer:

B)

Explanation:

I hope this help's :)

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A flute is designed so that it plays a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20
irinina [24]

Answer:

0.655 m

13.468°C

Explanation:

v = Speed of sound at 20.0°C = 343 m/s (general value)

For one both end open we have the expression

L=\dfrac{\lambda_1}{2}\\\Rightarrow L=\dfrac{\dfrac{v}{f_1}}{2}\\\Rightarrow L=\dfrac{\dfrac{343}{261.6}}{2}\\\Rightarrow L=0.655581039755\ m

The length of the flute is 0.655 m

Beat frequency is given by

\Delta f=f_1-f_2\\\Rightarrow 3=261.6-f_2\\\Rightarrow f_2=261.6-3\\\Rightarrow f_2=258.6\ Hz

Velocity of the wave is

v=f_2\lambda_1\\\Rightarrow v=258.6\times \dfrac{343}{261.6}\\\Rightarrow v=339.066513761\ m/s

The temperature is given by

T=273(\dfrac{v}{331})^2\\\Rightarrow T=273(\dfrac{339.066513761}{331})^2\\\Rightarrow T=286.468227799\ K=286.468227799-273=13.468227799^{\circ}C

The temperature of the room is 13.468°C

4 0
3 years ago
These graphs show measurements of the density of air as different sound waves pass a single point.
Ymorist [56]

Answer: D!

It is the option with the greatest amplitude.

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Review. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always st
anyanavicka [17]

To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,

The circumference of the planet is,

\phi = 25.1m

Under the mathematical value the radius would be

\phi = 2\pi r

r = \frac{25}{2\pi}

r = 3.9788km

Using second equation of motion

x = \frac{1}{2} at^2

Replacing the values given,

1.4 = \frac{1}{2} a (29.2)^2

Rearranging and solving for 'a' we have,

a = 0.003283m/s^2

Using the value of acceleration due to gravity from Newton's law we have that

a = \frac{GM}{r^2}

Here,

r = Radius of the planet

G = Gravitational Universal constant

M = Mass of the Planet

\frac{(6.67*10^{-11})*M}{(3.9788*10^3)^2} = 0.003283

M = 7.79201*10^{14}kg

Therefore the mass of this planet is 7.79201*10^{14}kg

5 0
4 years ago
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