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xeze [42]
3 years ago
13

Select the correct answer.

Physics
2 answers:
Vesna [10]3 years ago
8 0

Answer:

C

Explanation:

Radiant= list onto solar panels, Electric= solar into power, Radiant= Electric into light

ivanzaharov [21]3 years ago
3 0

Answer:

C.) radiant energy - electric energy - radiant energy

Explanation:

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Which of the following does not affect the apparent brightness of a star.
Zigmanuir [339]
A) how old it is :)))
3 0
3 years ago
Read 2 more answers
The following three resistors (Ohms) are in parallel with one voltage source. What is the total current running through circuit?
Zepler [3.9K]

Explanation:

The total resistance Rt is given by

1/Rt = 1/Ra + 1/Rb + 1/Rc = 1/5 + 1/4 + 1/7 = 0.593

or Rt = 1.69 ohms

The total current I running through the circuit is

I = V/Rt = 2.37 A

6 0
3 years ago
Calculate the KE of a car which has a mass of 1000 kg and is moving at the rate of 20 m/s
Elan Coil [88]

Answer:

The kinetic energy of the car is 2.0\cdot 10^5 J

Explanation:

3 0
2 years ago
Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin
Arte-miy333 [17]

Answer:

x=\dfrac{r}{\sqrt2}

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

E=K\dfrac{Q}{(r^2+x^2)^{3/2}}

For maximum condition

\dfrac{dE}{dx}=0

E=K{Q}{(r^2+x^2)^{-3/2}}

\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}

For maximum condition

\dfrac{dE}{dx}=0

K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0

r^2+x^2-3x^2=0

x=\dfrac{r}{\sqrt2}

At x=\dfrac{r}{\sqrt2} the electric field will be maximum.

3 0
2 years ago
The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an
Mrrafil [7]

Answer:

v_1  = 3.5 \ m/s

Explanation:

Given that :

mass of the SUV is  = 2140 kg

moment of inertia about G , i.e I_G = 875 kg.m²

We know from the conservation of angular momentum that:

H_1= H_2

mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2

2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2

1637.1 v_1 = 3841.575 \omega_2

\omega_2 = \frac{1637.1 v_1}{3841.575}

\omega _2 = 0.4626 \ v_1

From the conservation of energy as well;we have :

T_2 +V_{2  \to 3} = T_3 \\ \\ \\  \frac{1}{2} I_A \omega_2^2 - mgh =0

[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0

706.93 \ v_1^2 - 8657.49 =0

706.93 \ v_1^2  = 8657.49

v_1^2  =  \frac{8657.49}{706.93 }

v_1 ^2 =  12.25

v_1  = \sqrt{ 12.25

v_1  = 3.5 \ m/s

6 0
3 years ago
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