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s344n2d4d5 [400]
2 years ago
14

A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And

oil are 0.5kj/kg.k and 3.5kj/kg.k calculate the entropy change in steel,oil and in the universe.​
Physics
1 answer:
Naily [24]2 years ago
8 0

Answer:

700J

Explanation:

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A car is traveling in a race.The car went from initial velocity of 35m/s to the final velocity of 65m/s in 5 seconds what was th
I am Lyosha [343]
Acceleration is the change in velocity divided by time. The change in velocity is -30m/s and time is 5s. If you divide -30m/s by 5s, you get -6m/s<span>².</span>
8 0
2 years ago
Read 2 more answers
A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a
Pepsi [2]

Answer:

88.3

Explanation:

Emf in a rotating coil is given by rate of change of flux:

E= dФ/dt=(NABcos∅)/ dt

N: number of turns in the coil= 80

A: area of the coil= 0.25×0.40= 0.1

B: magnetic field strength= 1.1

Ф: angle of rotation= 90- 37= 53

dt= 0.06s

E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V

4 0
2 years ago
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Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
2 years ago
The weight of a bucket is 186 N. The bucket is being raised by two ropes. The free-body diagram shows the forces acting on the b
amm1812
Fnet=(115+106)-186= 34 N

mass=Force/g= 186N/9.8m/s^2 = 18.98 kg

a=fnet/mass => 34N/18.98kg = 1.79 m/s^2

so A= 1.8m/s^2
4 0
2 years ago
Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear
Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

8 0
2 years ago
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