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PSYCHO15rus [73]

3 years ago

15

Calculate the electrical energy expended in a device across which the circuit voltage drops by 50 volts in moving a charge of 6

coulombs.

2 answers:

solmaris [256]3 years ago

5 0

1 volt = 1 joule per coulomb 50 volts = 50 joules per coulomb 50 joules/coulomb times 6 coulombs = 300 joules

olasank [31]3 years ago

4 0

$I=\dfrac{Q}{t} \\ \\ \\ W=VIt=V\dfrac{Q}{t}t=VQ=50\times 6=300$

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2 years ago

Help wit these questions someone.

romanna [79]

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, $\frac{1}{Req} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...$

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}$

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<h3>Q8.</h3>

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

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7 0

2 years ago

What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?

Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, $q=6.4\times 10^{-6}\ C$

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV$

So, the required electric potential is 192 kV.

3 0

2 years ago

What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur

AveGali [126]

Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing $a=76m^2$

We know that density of air $d=1.29kg/m^3$

Speed at top surface $v_2=290m/sec$ and speed at bottom surface $v_1=150m/sec$

According to Bernoulli's principle force is given by

$F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N$

4 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0

2 years ago