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2 years ago

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Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe

r where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.

1 answer:

levacccp [35]2 years ago

4 0

Answer:

Obviously Lengthen... $T = 2\pi \sqrt{L/g}$ or $g = 4\pi ^{2} L/g$

Explanation:

As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...

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Which energy source is formed when organic matter is trapped underground without exposure to air or moisture?. . A. natural gas.

nikklg [1K]

Answer:

The correct answer is option B. coal

Explanation:

Coal is made of remains of organic material including trees and other vegetation which got trapped beneath the earth’s surface or at the bottom of the swamps. After burial below the ground the organic material was acted upon by the high temperature and pressure in the absence of air to form peat. Peat after further processing for a longer period of time converted into coal

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2 years ago

At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat

vlabodo [156]

Answer:

$20.3125$ kJ/mol

Explanation:

$P_{i}$ = initial vapor pressure = 45.77 mm Hg

$P_{f}$ = final vapor pressure = 193.1 mm Hg

$T_{i}$ = initial temperature = 213.1 K

$T_{f}$ = final temperature = 243.7 K

$H$ = Heat of vaporization

Using the equation

$ln\left ( \frac{P_{f}}{P_{i}} \right ) = \left ( \frac{-H}{R} \right )\left ( \frac{1}{T_{f}} - \frac{1}{T_{i}}\right)$

$ln\left ( \frac{193.1}{45.77} \right ) = \left ( \frac{-H}{8.314} \right )\left ( \frac{1}{243.7} - \frac{1}{213.1}\right)$

$H = 20312.5$ J/mol

$H = 20.3125$ kJ/mol

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2 years ago

Please re-explain the following phrases in terms of momentum

zheka24 [161]

A. An object in motion will remain in motion unless acted upon by an outside force : The momentum of an object is constant unless an outside force acts on the object.

B. Force is defined as mass times acceleration : the rate of change of the momentum of a particle is proportional to the force F acting on it, hence the force is equal to <span>mass times acceleration.

C. </span>For every action there is an equal and opposite reaction : <span>to every action force there is an equal and opposite reaction force. </span>

8 0

2 years ago

A cup of hot coffee initially at 95 degrees C cools to 80 degrees C in 5 min while sitting in a room of temperature 21 degrees C

Oksana_A [137]

Answer:

When the temperature of the coffee is 50 °C, the time will be 20.68 mins

Explanation:

Given;

The initial temperature of the coffee T₀ = 95 °C

The temperature of the room = 21°C

Let T be the temperature at time of cooling t in mins

According to Newton's law of cooling;

$\frac{dT}{dt} \alpha (T-21)\\\\\frac{dT}{dt} = k (T-21)\\\\\frac{dT}{T-21} = kdt\\\\\int\limits {\frac{dT}{T-21}} = \int\limits kdt\\\\Log(T-21) =kt + Logc \\\\Log (\frac{T-21}{c} ) = kt\\\\T -21 = ce^{kt}\\\\At \ t = 0, T = 95\\\\95-21 = ce^0\\\\74 = c\\\\New, equation: T -21 = 74e^{kt}\\\\Again; when \ t= 5\ min, T = 80\\\\80 -21 = 74e^{5k}\\\\59 = 74e^{5k}\\\\e^{5k} = \frac{59}{74}\\\\ 5k = ln(\frac{59}{74})\\\\5k = -0.2265\\\\k = -0.0453$

When the temperature is 50 °C, the time t in min is calculated as;

$T -21 = 74e^{-0.0453t}\\\\50 -21 = 74e^{-0.0453t}\\\\29 = 74e^{-0.0453t}\\\\\frac{29}{74} = e^{-0.0453t}\\\\0.39189 = e^{-0.0453t}\\\\ln(0.39189 ) = {-0.0453t}\\\\-0.93677 = {-0.0453t}\\\\t = \frac{-0.93677}{-0.0453}\\\\ t = 20.68 \ mins$

Therefore, when the temperature of the coffee is 50 °C, the time will be 20.68 mins

4 0

2 years ago

A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

$\epsilon = -\frac{\Delta \Phi_B}{\Delta t}$ (1)

where

$\Delta \Phi_B$ is the variation of magnetic flux through the coil

$\Delta t = 2.0 ms = 0.002 s$ is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

$\Phi_B = BA$

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

$A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2$

At the beginning, the magnetic field is

$B_i = 1.5 T$

so the flux is

$\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb$

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

$\Phi_f = 0$

so the variation of magnetic flux is

$\Delta \Phi = 0-0.068 Wb=-0.068 Wb$

And substituting into (1) we find the average emf in the coil

$\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V$

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

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2 years ago