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2 years ago

List two ways of recording or measuring absolute time

Physics

1 answer:

enyata [817]2 years ago

7 0

Answer:

Stop watch

Stop clock

Explanation:

Absolute time is time measured in definite periods such as minutes, days, and years. It can be measured uising stop watch or stop clock

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Explanation:

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What is an expression for the vertical component of this vector?

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The correct choice is D .

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2 years ago

Which of the following objects has kinetic energy?

kotykmax [81]

Answer:

Explanation:

A body has kinetic energy that is moving

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2 years ago

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Ideally, when a thermometer is used to measure the temperature of an object, the temperature of the object itself should not cha

nordsb [41]

Answer : The temperature of water is, $57.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of thermometer = $804J/kg.^oC=0.804J/g^oC$

$c_2$ = specific heat of water = $4.18J/g.^oC$

$m_1$ = mass of thermometer = 33.0 g

$m_2$ = mass of water = 149 g

$T_f$ = final temperature = $56.0^oC$

$T_1$ = initial temperature of thermometer = $15.4^oC$

$T_2$ = initial temperature of water = ?

Now put all the given values in the above formula, we get:

$33.0g\times 0.804J/g^oC\times (56.0-15.4)^oC=-149g\times 4.18J/g.^oC\times (56.0-T_2)^oC$

$T_2=57.7^oC$

Therefore, the temperature of water is, $57.7^oC$

5 0

2 years ago

Electric power is to be generated by installing a hydraulic turbine-generator at a site 70 m below the free surface of a large w

Tasya [4]

Answer:

$\eta_{turbine} = 0.777 = 77.7\%$

$\eta_{combined} = 0.728 = 72.8\%$

Explanation:

First we calculate the power input to the turbine. The input power will be equal to the potential energy of water per unit time:

Input Power = $P_{in} = \frac{Work}{Time} = \frac{Potential\ Energy\ of\ Water}{t} \\P_{in} = \frac{(mass)(g)(height)}{Time} = (mass flow rate)(g)(height)\\\\P_{in} = (1500\ kg/s)(9.81\ m/s^2)(70\ m)\\P_{in} = 1.03\ x\ 10^6\ W = 1030\ KW$

Now, for turbine efficiency:

$\eta_{turbine} = \frac{Mechanical\ Power\ Out}{P_{in}}\\\\\eta_{turbine} = \frac{800\ KW}{1030\ KW}\\\\\eta_{turbine} = 0.777 = 77.7\%$

for generator efficiency:

$\eta_{generator} = \frac{Power\ Generation}{Mechanical\ Power\ Out}\\\\ \eta_{generator} = \frac{750\ KW}{800\ KW}\\\\\eta_{turbine} = 0.9375 = 93.75\%$

Now, for combined efficiency:

$\eta_{combined} = \eta_{turbine}\ \eta_{generator}\\\\\eta_{combined} = (0.777)(0.937)\\\eta_{combined} = 0.728 = 72.8\%$

4 0

2 years ago