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olasank [31]
3 years ago
5

3. If the potential difference is 120 volts and the resistance is 60 ohms, what is the current?

Physics
1 answer:
Ede4ka [16]3 years ago
8 0

Answer:

2 amperes

Explanation:

According to Ohms law

v = ir

v = voltage

I = current

r = resistance

From the question

v = 120volts

r = 60 ohms

I = ?

Using ohms law ,

v = ir

120 = I x 60

Divide both sides by 60

120/60 = I x 60/60

2 = I

I = 2 amperes

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Precisely, water has to absorb 4,184 Joules of heat (1 calorie) for the temperature of one kilogram of water to increase 1°C. For comparison sake, it only takes 385 Joules of heat to raise 1 kilogram of copper 1°C.

Explanation:

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3 years ago
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Alex_Xolod [135]

Answer:

The proton has much greater mass

Explanation:

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2 years ago
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A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?
GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

3 0
3 years ago
The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below. The vertical axis is the maximum
Rudiy27

Answer:

Check the explanation

Explanation:

A) 7th Harmonic. (Of an open ended pipe, odd harmonics are allowed (3rd overtone))

b) f = n v / 4 L

n = 7

f = 7 x 350 / 4 x 0.41 = 1493.9 Hz

c) Let level of water H, If reduces the effective length of pipe

Using, f = n v / 4 Leff

n = 1

251.8 = 1 x 350 / 4 ( 0.41 - H)

H = 0.0625m

H = 6.25 cm

5 0
3 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
3 years ago
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