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Andrej [43]
3 years ago
13

Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. T

hen the rate of heat transfer to the water is 225.7 kJ/min 45.1 kJ/min 53.5 kJ/min 0.84 kJ/min 41.8 kJ/min
Physics
1 answer:
diamong [38]3 years ago
6 0

Answer:

Rate of heat is 45.1 kJ/min

Explanation:

Heat required to evaporate the water is given by

Q = mL

here we know that

L = 2.25 \times 10^6 J/kg

now we have

Q = (0.200)(2.25 \times 10^6 J/kg)

Q = 452.1 kJ

now the power is defined as rate of energy

P = \frac{Q}{t}

P = \frac{452.1 kJ}{10}

P = 45.1 kJ/min

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The answer is A. Metals
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Que distancia se desplaza un objeto que se mueve con velocidad de 72km/h durante 10 min?
Allushta [10]

Answer:

I would love to help, Could you put the question in English?

Explanation:

4 0
3 years ago
An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto
Lemur [1.5K]

Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

R\propto\dfrac{1}{D}

The power is inversely proportional to the resistance.

P\propto\dfrac{1}{R}

P\propto D^2

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}

Put the value into the formula

\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}

\dfrac{D_{i}}{D_{f}}=0.95

D_{i}=0.95 D_{f}

Hence, The factor of the diameter is 0.95.

7 0
3 years ago
Read 2 more answers
a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r
dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

#SPJ4

4 0
2 years ago
A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
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