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igor_vitrenko [27]

1 year ago

7

Another term for electromotive force is _____. voltage current resistance power

1 answer:

Ray Of Light [21]1 year ago

3 0

Voltage because I said so

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A compressed spring does not have elastic potential energy.

Alexus [3.1K]

Answer:

False

Explanation:

The second you let go its gonna release kinetic energy that's why it's potential

6 0

1 year ago

A 4.0 Ω resistor, an 8.0 Ω resistor, and a 12.0 Ω resistor are connected in parallel across a 24.0 V battery. What is the equiva

dsp73

PART A)

Equivalent resistance in parallel is given as

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

now we have

$\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12}$

$R = 2.18 ohm$

PART B)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

$V = iR$

for 4 ohm resistance we have

$24 = 4(i)$

$i = 6 A$

PART C)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

$V = iR$

for 8 ohm resistance we have

$24 = 8(i)$

$i = 3 A$

3 0

2 years ago

Read 2 more answers

A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined r

Elena-2011 [213]

Answer: $A=23.57^{\circ}$

Explanation:

Given

velocity=100 ft/s

height of landing zone=10 ft

Equation of $x(t)=100 t\cos (A)$

$y(t)=-16t^2+100t\sin (A)+10$

Maximum height=35 feet

at maximum height

$\frac{\mathrm{d} Y}{\mathrm{d} t}=0$

$\frac{\mathrm{d} x}{\mathrm{d} t}=-32t+100\sin A$

$t=3.125\sin A$

At $t=3.125\sin A$

$Y(t)=35=-16\times (3.125\sin A)^2+100\times 3.125\sin A\times \sin A+10$

$312.5\sin^2 A-156.25\sin^2 A=25$

$sin^2 A=0.16$

$A=23.57^{\circ}$

8 0

1 year ago

Nitrogen (n2) gas within a piston–cylinder assembly undergoes a compression from p1 = 20 bar, v1 = 0.5 m3 to a state where v2 =

Bingel [31]

Part a)

As we know that

$P_1V_1^{1.35} = P_2V_2^{1.35}$

here we know that

P1 = 20 bar

V1 = 0.5 m^3

V2 = 2.75 m^3

from above equation

$20* 0.5^{1.35} = P * (2.75)^{1.35}$

$P = 2 bar$

so final state pressure will be 2 bar

Part b)

now in order to find the work done

$W = \int PdV$

$W = \int \frac{c}{V^{1.35}}dV$

$W = c\frac{V^{-0.35}}{-0.35}$

$W = \frac{P_1V_1 - P_2V_2}{0.35}$

$W = \frac{20* 0.5 - 2 * 2.75}{0.35}* 10^5 = 12.86 * 10^5 J$

3 0

1 year ago

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Which of these is the largest? <br> a. star<br> b. nebula<br> c. galaxy<br> d. sun

Keith_Richards [23]

I would say it’s between b and c

5 0

1 year ago

Read 2 more answers