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miss Akunina [59]
2 years ago
9

a water heater has a power rating of 1 kW. how many seconds will this heater take to boil 1 liter of water?

Physics
1 answer:
Tanzania [10]2 years ago
5 0
Answer:
4.65 minutes.

Explanation:
Theoretical Assumptions (not realistic): 1) ALL of the heat energy is transferred to the water, and 2)the water is uniformly heated. 3) The water density is 1.0g/mL.
Because we are only heating it TO the “boiling point”, ONLY the energy to heat the water from 20 to 100’C using the heat capacity or specific heat of water is required. A “Watt” is 1 Joule/second. The specific heat of water is 4.184 J/g-’C. Water mass is 1000g. The energy required to heat it is therefore
1000
g
⋅
4.184
(
J
g
’
C
)
⋅
(
100
−
20
)
’
C
=
334720
J

The heater supplies 1200 J/s, so the time to provide this heat to the water is
334720
J
1200
(
J
s
)
=
278.9
s
or 4.65 minutes.
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Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many
Vlada [557]

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

8 0
2 years ago
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Phantasy [73]

Answer:

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block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th
Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0
2 years ago
Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750
lesya692 [45]

Answer:

Part a)

f = 1911.5 Hz

Part b)

\lambda = 0.186 m

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

f' = f_0 (\frac{v + v_o}{v - v_s})

here we know that

f_0 = 1750 Hz

v_o = 15 m/s

v_s = 15 m/s

now we will have

f' = (1750)(\frac{340 + 15}{340 - 15})

f' = 1911.5 Hz

Part b)

Apparent wavelength is given by the formula

\lambda = \frac{v_{relative}}{f_{app}}

here we will have

\lambda = \frac{340 + 15}{1911.5}

\lambda = 0.186 m

3 0
2 years ago
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