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Molodets [167]
3 years ago
12

As the temperature of an object rises, so does the

Physics
2 answers:
FrozenT [24]3 years ago
5 0

Answer:

The correct answer Is C! Thermal energy of the object

Explanation:

nikklg [1K]3 years ago
3 0
C. The object is not in motion, ruling out A. We are not adding mass in any way, nor does adding heat to object increase its mass, therefore also ruling out B. Finally, we are not changing the object's position in such a way that gives it a higher ability to do work, ruling out D. 
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If you do 1500 J of work hoisting a 20 kg bale of hay , to what height did you lift it
strojnjashka [21]

Explanation:

W = PE

W = mgh

1500 J = (20 kg) (9.8 m/s²) h

h = 7.65 m

Round as needed.

4 0
3 years ago
in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at
Tema [17]
<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

</span>

3 0
3 years ago
What does Kepler's first law of planetary motion imply?
BlackZzzverrR [31]

<u>Answer:</u>

The correct answer option is D.  The distance between the planet and the Sun changes as the planet orbits the sun.

<u>Explanation:</u>

Kepler’s laws of planetary motion, derived by the German astronomer Johannes Kepler, are the laws of physics that describe the motions of the planets in the solar system.

According to the Kepler's first law of planetary motion: the path on which the planets orbit around the sun is elliptical in shape, with the center of the sun at one focus.

Therefore, the distance between the Sun and the planets vary as the planet orbit around the sun.

6 0
3 years ago
The hottest climates on Earth are located near the Equator because this region
Ber [7]

You should select Choice-4 .

4 0
3 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
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