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3 years ago

The Earth orbits around the sun because the gravitational force that the sun

Physics

1 answer:

kotykmax [81]3 years ago

5 0

<h3>Question -:</h3>

The Earth orbits around the sun because the gravitational force that the sun

exerts on the Earth:

O A. causes Earth's acceleration toward the sun.

O B. is very small because the sun is so far from the Earth.

O c. is smaller than the force the Earth exerts on the sun.

O D. pushes the Earth away from the sun.

<h3>Answer -:</h3>

O A. causes Earth's acceleration toward the sun.

I hope this helps, have a nice time ahead!

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A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an

Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

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The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum

kirill115 [55]

Answer:

The maximum voltage is 41.92 V.

Explanation:

Given that,

Peak voltage = 590 volts

Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is $1.20×10^{-2}\ \mu F$ .

We need to calculate the resonance frequency

Using formula of frequency

$f=\dfrac{1}{2\pi\sqrt{LC}}$

Put the value into the formula

$f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}$

$f=2356.88\ Hz$

We need to calculate the maximum current

Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi\times f\times C\times V_{c}$

$I=2\pi\times2356.88\times1.20\times10^{-8}\times590$

$I=0.1048\ A$

Impedance of the circuit is

$z=\sqrt{R^2+(X_{L}^2-X_{C}^2)}$

At resonance frequency $X_{L}=X_{C}$

$Z=R$

We need to calculate the maximum voltage

Using ohm's law

$V=I\times R$

$V=0.1048\times400$

$V=41.92\ V$

Hence, The maximum voltage is 41.92 V.

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