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xz_007 [3.2K]
3 years ago
7

A grandfather clock depends on the period of a pendulum to keep correct time. Suppose a grandfather clock is calibrated correctl

y and then a mischievous child slides the bob of the pendulum downward on the oscillating rod. The grandfather clock will run:
Physics
2 answers:
sweet-ann [11.9K]3 years ago
7 0

The grandfather clock will run: slower

<h3>Explanation: </h3>

A grandfather clock depends on the period of a pendulum to keep correct time. Suppose a grandfather clock is calibrated correctly and then a mischievous child slides the bob of the pendulum downward on the oscillating rod.

A pendulum is a weight that get suspended from a pivot, therefore it can swing freely. A pendulum clock is the clock using a pendulum, a swinging weight, as its timekeeping element.

Calibrate in physics means to ascertain the caliber of and to determine, rectify, or mark the graduations of something, such as a thermometer tube.

A bob is the weight located on the end of a pendulum found in pendulum clocks.

The time period of simple pendulum is given by:

T = 2 \pi \sqrt{\frac{L}{g} }

The time period of a simple pendulum is defined as the time taken by the pendulum to finish one full oscillation.

Where l is the effective length of the pendulum, and g is the acceleration due to gravity

The grandfather clock will run: Slower. Because as the length of the pendulum increase, the tie period of the pendulum also increase.  It means the clock run slow.

Learn more about   pendulum brainly.com/question/8411900

#LearnWithBrainly

Setler [38]3 years ago
6 0

Answer:

Slow

Explanation:

As we know that the time period of simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

Where, l be the effective length of the pendulum and g be the acceleration due to gravity

As the length of the pendulum increase, then the tie period of the pendulum will also increase.

It means the clock run slow.

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c

Explanation:

though c is wider it has more water.

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3 years ago
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1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin
aleksandr82 [10.1K]

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

         W = ΔK

In that case as the car stops v_f = 0

the work is

          W = -fr d

we substitute

          - fr d₀ = 0 - ½ m v₀²

           d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

          v = 2 v₀

using the same expressions we find

           d = ½ m (2v₀)² / fr

           d = 4 (½ m v₀² / fr)

           d = 4 d₀o

3 0
3 years ago
A motorcycle running on gasoline wastes a large amount of energy mainly as A) heat energy and sound energy. B) light energy and
vlada-n [284]

A motorcycle mainly wastes energy as heat <u>energy</u> and <u>sound</u> energy. In the engine, chemical energy is transformed into mechanical energy. However, the engine is inefficient and much of the chemical energy is lost as heat energy. Also, some of the energy is transformed to sound energy. This explains why the motorcycle is noisy and has an exhaust pipe.

3 0
4 years ago
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
HELP
Sliva [168]

The stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.

<h3>What is kinetic energy?</h3>

Kinetic energy is the energy of a body possessed due to motion.

This means that for an object to possess kinetic energy, it must be in motion.

The kinetic energy is measured in Joules, which is a product of the mass of the substance and the time taken to travel a distance.

A stopwatch is an instrument used to measure time as one of the components of kinetic energy.

Therefore, the stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.

Learn more about kinetic energy at: brainly.com/question/12669551

8 0
2 years ago
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