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Natalija [7]
3 years ago
11

The initial temperature of 150 g of ethanol was 22 C. What will be the final temperature of the ethanol if 3240 J was needed to

raise the temperature of the ethanol
Physics
1 answer:
AfilCa [17]3 years ago
8 0

Answer:

<h3>30.405°C</h3>

Explanation:

Using the formula for heart capacity;

Q = mcΔt

m is the mass = 150g = 0.15kg

initial temperature = 22°C

Quantity of heat = 3240J = 3.24kJ

specific heat capacity of ethanol = 2.57 [kJ/kg K]

Substitute and get the final temperature

3.240 = 0.15(2.57)(T - 22)

3.240 = 0.3855(T-22)

3.240/0.3855 = T - 22

8.405 = T - 22

T = 22+8.405

T = 30.405°C

Hence the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol is 30.405°C

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Answer:

6.25

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Explanation:

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5. A cheetah with a mass of 70 kg was clocked running at 72 mph (32 m's). How many joules of
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Answer:

Kinetic energy = 35840 Joules

Explanation:

Given the following data;

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To find the kinetic energy;

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

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K.E = \frac{1}{2}MV^{2}

Substituting into the equation, we have;

K.E = \frac{1}{2}*70*32^{2}

K.E = \frac{1}{2}*70*1024

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Therefore, the kinetic energy possessed by the cheetah is 35840 Joules.

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3 years ago
A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9
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Answer:

C = 46.891 \mu F

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

s = \frac{P}{P.F} < COS^{-1} 0.65

S = \frac{1250}{0.65} < 49.45

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]

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Q_{new} = P tan [cos^{-1} P.F new]

Q_{new} = 1250 [tan[cos^{-1}0.9]]

Q_{new} = 605.40 VARS

Q_C = Q - Q_{new}

Q_C = 1461 - 605.4 = 855.6 vars

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C = \frac{Q_C}{ v_{rms}^2 \omega}

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3 years ago
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A protester carries his sign of protest, starting from the origin of an xyz coordinate system, with the xy plane horizontal. He
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Answer:

Part A:

Displacement\ vector= -40\hat i+20\hat j+25\hat k

Part B:

Magnitude of displacement=44.721359 m

Explanation:

Given Data:

40 m in negative direction of the x axis

20 m perpendicular path to his left (considering +ve y direction)

25 m up a tower (Considering +ve z direction)

Required:

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Solution:

Part A:

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where:

i,j,k are unit vectors

Part B:

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Displacement= -40\hat i+20\hat j+0\hat k

Magnitude of displacement:

Magnitude\ of\ displacement=\sqrt{(-40)^2+(20)^2+0^2} \\Magnitude\ of\ displacement=44.721359\ m

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