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svetoff [14.1K]
3 years ago
5

Which is a front in which cold air is replacing warm air at the surface?

Physics
2 answers:
Galina-37 [17]3 years ago
6 0
The answer should be a. cold front! a stationary front is when a cold front and warm front meet and do not move. STATIONARY IS INCORRECT!
Snezhnost [94]3 years ago
5 0

Answer:

C.stationarty front

Explanation:

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A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same ma
Alex

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}

in this case:

3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s

the initital kinetic energy is:

K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ

and the final:

K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ

The energy lost is given by:

E_l=|K_f-K_i|\\E_l=23kJ

8 0
3 years ago
Good night all of you​
Helen [10]

Answer: Good night

Explanation:

3 0
3 years ago
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What happens when a wave reflects?
solong [7]
They bounce off just like sun rays when they heat the atmosphere they bounce off reflection
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4 years ago
If you applied a 11 N force to a box and made it travel 108 m in 7
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Explanation:

due to the local people in the advantage and the value of that house so that there were no people this answer was belonging to define in refinery of the circuit

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4 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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