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Harlamova29_29 [7]
3 years ago
7

Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat

ion due to gravity is 1.63 m/s2 .
Physics
1 answer:
vichka [17]3 years ago
3 0
The period of a pendulum is given by
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
\frac{T_m}{T_e} =  \frac{2 \pi  \sqrt{ \frac{L}{g_m} } }{2 \pi  \sqrt{ \frac{L}{g_e} } }=    \sqrt{ \frac{g_e}{g_m} }
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is g_e = 9.81 m/s^2 while on the Moon is g_m=1.63 m/s^2, the ratio between the period on the Moon and on Earth is
\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45

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A technician wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition
Nana76 [90]

Answer:

λ = 5.656 x 10⁻⁷ m = 565.6 nm

Explanation:

Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:

\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x\ d}{L}

where,

λ = wavelength = ?

Δx = fringe spacing = 1.6 cm = 0.016 m

L = Distance between slits and screen = 4.95 m

d = slit separation = 0.175 mm = 0.000175 m

Therefore,

\lambda = \frac{(0.016\ m)(0.000175\ m)}{4.95\ m}\\\\

<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>

6 0
3 years ago
If the person drops box from 3.8 m how much energy is transferred from potential energy to kinetic energy
kotykmax [81]

Answer:

Kinetic energy

When work is done the energy is transferred from one type to another. This transferred energy may appear as kinetic energy.

For example, when you pedal your bicycle so that its speed increases, you are doing work to transfer chemical energy from your muscles to the kinetic energy of the bicycle.

Kinetic energy is the energy an object possesses by virtue of its movement. The amount of kinetic energy possessed by a moving object depends on the mass of the object and its speed. The greater the mass and the speed of the object the greater its kinetic energy.

The kinetic energy Ek of an object of mass m at a speed v is given by the relationship

{E_k} = \frac{1}{2}m{v^2}

m is the mass of the object in kilograms ( kg) and v is the speed of the object in metres per second ( m\,s^{-1}).

Explanation:

When work is done on an object it may also lead to energy being transferred to the object in the form of gravitational potential energy of the object.

Gravitational potential energy is the energy an object has by virtue of its position above the surface of the Earth. When an object is lifted, work is done. When work is done in raising the height of an object, energy is transferred as a gain in the gravitational potential energy of the object.

For example, suppose you lift a suitcase of mass m through a height h. The weight W of the suit case is a downward force of size mg. In lifting the suitcase, you would have to pull upwards on it with a force equal in size to its weight, mg.

Two suitcases. One has a green force arrow pointing up labelled F and a purple force arrow pointing down labelled 'Weight = mg'. The other case is raised by a height labelled h.

Suitcases with forces and height labelled

When this force (equal to the weight mg, but upwards) is applied to the suitcase over the distance h:

Work\,done=force\,\times\,distance\,upwards=mg\,\times\,h

This energy is transferred to potential energy when raising the object through a known height.

Energy = mass \times gravitational\,field\,strength \times height

E = m \times g \times h

This is the relationship used to calculate gravitational potential energy.

{E_p} = mgh

where m is the mass of the object in kilograms (kg), g is the gravitational field strength, (for positions near the surface of the Earth g = 9∙8 newtons per kilogram ( N kg ^{-1} and h is the height above the surface of the Earth in metres ( m).

8 0
3 years ago
Consider Newton's Law of Universal Gravitation: FG= G (m1 m2)/d2 .
bixtya [17]

Answer:

C

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

Explanation:

The formula for the force of gravity between two objects is

F=G\frac{m_1 m_2}{d^2}

where

G is the gravitational constant

m1, m2 are the masses of the two objects

d is the separation between the two objects

We notice the  following:

- F is directly proportional to the masses, F\propto m_1, m_2. This means that if one of the masses increases, then the force between them, F, increases in a proportional way

- F is inversely proportional to the square of the distance, F\propto \frac{1}{d^2}. This means that if the distance between the two objects is increased, the force between them will decrease, and vice-versa.

So, the correct answer is

C

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

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