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makkiz [27]
3 years ago
6

Please help on this one?

Physics
1 answer:
kupik [55]3 years ago
7 0

Answer: A)30V. First find the current of the circuit. I=V/R(total resistance). So I=60/120=0.5. Now to find voltage drop in R3 use ohms law as given. V(of 3)=(0.5)(60)=30V

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Which of the following is not part of a circuit? <br>film <br>Load<br> Key <br>Cell
qaws [65]

Explanation:

I believe <u>f</u><u>i</u><u>l</u><u>m</u><u> </u><u>i</u><u>s</u><u> </u>not part of a circuit

8 0
3 years ago
What is the internal energy of 3.00 mol of N2 gas at 25 degrees C? To solve this problem, use the equation: U electronic gas = 5
allochka39001 [22]

<u>Given data</u>

Determine Internal energy of gas N₂,  (U) = ?

Temperature (T) = 25° C

                          = 25+273 = 298 K,

Gas constant (R) = 8.31 J/ mol-K ,

Number of moles (n) = 3 moles,

<u>Internal energy of N₂ </u>

Internal energy is a property of thermodynamics, the concept of internal energy can be understand by ideal gas. For example N₂, the observations for oxygen and nitrogen at atmospheric temperatures,  f=5,  (where f is translational  degrees of freedom).

       So per kilogram of gas,

          The internal energy (U) = 5/2 .n.R.T

                                                  = (5/2) × 3 × 8.31 ×298

                                                  = 18572.85 J

<em>The internal energy of the N₂ is 18,572.85 J and it is approximately equal to 18,600 J given in the option B.</em>

 



8 0
3 years ago
Read 2 more answers
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length t
gladu [14]

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

$t=\sqrt{\frac{2H}{g}}$

v_1t= 0.35  ...........(i)

v_0t= 1.09  ...........(ii)

Dividing the equation (ii) by (i)

$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$

∴ v_0=3.11  \ v_1

Now loss of energy  = change in the kinetic energy

$W=\frac{1}{2} m [v_0^2-v_1^2]$

$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$

$W=7.307\times 10^{-3} \ v_0^2$

If f is average friction force, then

(f)(L) = W

(f) (1) = $7.307\times 10^{-3} \ v_0^2$

(f)  = $7.307\times 10^{-3} \ v_0^2$

3 0
3 years ago
A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva
AleksandrR [38]

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

5 0
4 years ago
The atmospheric features of Neptune are easier to see than those of Uranus because A. Neptune has greater warmth and less haze.
Lelechka [254]
If I’m being honest I think it’s a or b
4 0
3 years ago
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