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11Alexandr11 [23.1K]
2 years ago
10

Please help me out !!

Physics
1 answer:
blagie [28]2 years ago
4 0
I think the answer would be letter a
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An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric
ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0
3 years ago
If you throw a tennis ball in the air,what type of motion does it have?
Alenkasestr [34]

-- Accelerated motion. Its speed changes constantly. If it's not tossed STRAIGHT up, its direction also changes constantly. Even if it IS tossed straight up, its direction changes when it gets to the top and starts falling.

-- Projectile motion. There is vertical force on the object because of gravity, but no horizontal force.

5 0
3 years ago
Approximate the field by treating the disk as a +6.1 C point charge at a distance of 21 cm Answer in units of N/C.
DaniilM [7]

Answer:

Electric field, E=1.24\times 10^{12}\ N/C

Explanation:

It is given that,

Charge, Q = +6.1 C

Distance, r = 21 cm = 0.21 m

We need to find the electric field. It is given by :

E=k\dfrac{Q}{r^2}

E=9\times 10^9\times \dfrac{6.1}{(0.21)^2}

E=1.24\times 10^{12}\ N/C

So, the electric field at this distance is 1.24\times 10^{12}\ N/C. Hence, this is the required solution.

7 0
2 years ago
How much smaller is an atom than a speck of dust
IceJOKER [234]

Answer:

an atom is about a trillion times smaller then a speck of dust.

Explanation:

hope this helps :)

3 0
3 years ago
The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at le
Vinil7 [7]

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

Then if the shuttle starts from rest, we have

S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

S = distance traveled, m

a = acceleration of the motion, m/s²

t = time of travel

S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far

3 0
3 years ago
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