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Rufina [12.5K]
3 years ago
10

Imagine we cook an egg by immersing it into water which is boiled by an electric heater. The heater utilizes a current I =10 A a

t a voltage V = 120 volts for a time t = 120 s lf the change in energy of the newly-cooked egg over its raw torm is given by ΔEngg = 28.8 kJ, what is the amount of energy wasted in the process?
Physics
1 answer:
tatuchka [14]3 years ago
8 0

Answer:

Q' = 115.2 KJ

Explanation:

Given that

Current I = 10 A

Voltage = 120 V

Time t= 120 s

The energy supplied Q= V I t

Q= 10 x 120 x 120 J

Q= 144 x 1000 J

Q= 144 KJ

The change in the energy ΔEngg = 28.8 kJ

By using energy conservation

Q= Q' +  ΔEngg

Q'=Wasted energy

Now by putting the values in the above equation

144 = Q' + 28.8 KJ

Q' = 115.2 KJ

Therefore the waste energy will be 115 kJ.

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murzikaleks [220]
I think it might be C :)
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3 years ago
What is the magnetic flux linkage, in units of Weber, for a coil of 360 turns and cross sectional area of 0.133 m^2 when the mag
zalisa [80]

Answer:

83.3 Wb

Explanation:

The magnetic flux linkage through the coil is given by:

N\phi = BAN sin \theta

where

B is the magnetic field strength

A is the cross sectional area

N is the number of turns in the coil

\theta is the angle between the direction of the field and the normal to the coil

In this problem:

B = 1.74 T

A = 0.133 m^2

N = 360

\theta=90^{\circ}

Therefore, the magnetic flux linkage is

N\phi = (1.74 T)(0.133 m^2)(360) sin 90^{\circ}=83.3 Wb

7 0
3 years ago
What is the kinetic energy of a hammer that starts from rest and decreases its potential energy by 10 kJ?
erma4kov [3.2K]

Answer:

final kinetic energy of the hammer is 10 kJ

Explanation:

As we know that there is no non conservative force on the system

So here we can use the theory of mechanical energy conservation

So we will have

\Delta K + \Delta U = 0

here we know that

\Delta U = - 10 kJ

from above expression now

K_f - K_i - 10 kJ = 0

K_f - 0 = 10 kJ

so final kinetic energy of the hammer is 10 kJ

3 0
3 years ago
Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy di
dlinn [17]

Answer:

E=29\times 10^{-5}eV

Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

E^{+}=E+\mu B,

E^{-}=E-\mu B,

E^{0}=E

Here, E is the energy in the absence of electric field.

And

E^{+} and E^{-} are the highest and the lowest energies.

The difference of these energies

\Delta E=2\mu B

\mu=9.3\times 10^{-24}J/T is known as Bohr's magneton.

B=2.5 T,

Therefore,

\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J

Now,

Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is E=29\times 10^{-5}eV

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3 years ago
(15 POINTS) Victor drew a diagram to show the life cycle of a low-mass star.
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The answer is c so the area can grow
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