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lana [24]
3 years ago
6

Suppose a pendulum clock has been calibrated to be accurate in San Francisco, where g = 9.800 m/s2 . In Camrose, g = 9.811 m/s2

is slightly larger due to the effect of Earth’s rotation at a higher latitude. Explain why the clock will either run perfectly, run too quickly, or run too slowly in Camrose.
Physics
1 answer:
tamaranim1 [39]3 years ago
8 0

The clock would run too slowly in Camrose.

Since the period of the pendulum T = 2π√(L/g) where L = length of clock and g = acceleration due to gravity.

Now since g = 9.800 m/s² in San Francisco and g = 9.811 m/s² in Camrose, we see that g increases.

From the expression for the period T, since L is constant, we find that

T ∝ 1/√g

Since g increases, so, T would decrease.

Thus, the clock would run too slowly in Camrose.

Learn more about pendulum clock here:

brainly.com/question/12405819

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Using 15 percent as machine energy efficiency, what is the actual work output if the total work input is 7500 kilojoules​?
g100num [7]

Answer:

The actual work output is 1,125 kilojoules.

Explanation:

Energy efficiency is defined as the efficient use of energy. An appliance, process, or facility is energy efficient when it consumes less than the average amount of energy to perform an activity. Then, energy efficiency is the ratio between the amount of energy used in an activity and the amount expected to be carried out.

Efficiency is calculated as:

efficiency = output / input

Where output is the amount of mechanical work (in watts) or energy consumed by the process (in joules), and input (input) is the amount of work or energy that is used as input to carry out the process.

In this case:

  • Efficiency always has a value between 0 and 1. In this case, efficiency=0.15
  • output=?
  • input= 7500 kilojoules

Replacing:

0.15=output/7500 kilojoules

Solving:

Output=0.15* 7500 kilojoules

output=1,125 kilojoules

<u><em>The actual work output is 1,125 kilojoules.</em></u>

7 0
3 years ago
What SI unit is used to measure mechanical energy
VashaNatasha [74]

The SI unit of energy is the Joule .

Any kind of energy ... electrical, mechanical, nuclear, solar,
wind etc.  It's easy to change from one form to another ... we
do it every day ... and they all have the same unit.

7 0
4 years ago
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Explanation:

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3 years ago
The charge of an electron is -1.60x10-19 C. A current of 1 A flows in a wire carried by electrons. How many electrons pass throu
faltersainse [42]

Answer: 6.241\times 10^{18} electrons pass through a cross section of the wire each second.

Explanation:

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

Given : Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

To calculate the charge passed we use the equation:

I=\frac{q}{t}

where,

I = current passed = 1 A

q = total charge = ?

t = time required = 1 sec

Putting values in above equation, we get:

1A=\frac{q}{1s}\\\\q=1A\times 1s=1C

When 96500C of electricity is passed , the electrons passed = 6.022\times 10^{23}

1 C of electricity is passed , the electrons passed = \frac{6.022\times 10^{23}}{96500}\times 1C=6.241\times 10^{18}

Hence, 6.241\times 10^{18} electrons pass through a cross section of the wire each second.

4 0
3 years ago
An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)
weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767

T₂ = 290.15 × 3.17767 = 922.00139

\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3}  = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12

\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702

Therefore;

T_4 = \frac{1383.002}{2.702} =511.859 \ k

Q_1 = c_p(T_3-T_2)

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

c_v(T_4-T_1)  = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1  \right )r_{v}^{k-1}}

Where:

β = Cut off ratio

Plugging in the values, we get;

\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1  \right )18^{1.4-1}}= 0.5191

Therefore;

\eta = \frac{\sum Q}{Q_1}

\therefore 0.5191 = \frac{150}{Q_1}

Heat supplied = \frac{150}{0.5191}  = 288.978 \ hp

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0
4 years ago
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