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3 years ago

State characteristics of the images formed by pin - hole camera

Physics

1 answer:

Leokris [45]3 years ago

8 0

Answer:

Real image

Explanation:

The picture is real, but it is reversed and tiny. An picture generated by a pinhole camera has certain features. As compared item, the image created by a pinhole camera is normally pretty small and looks reversed both on the vertically and horizontally axis.

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An astronaut on Pluto attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 10 complet

pickupchik [31]

Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

$T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2$

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3 years ago

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48

cluponka [151]

Answer:

The thickness is $\Delta y = 2.4 *10^{-6} \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 480 \ nm = 480*10^{-9} \ m$

The first order of the dark fringe is $m_1 = 16$

The second order of dark fringe considered is $m_2 = 6$

Generally the condition for destructive interference is mathematically represented as

$y = \frac{m \lambda}{2}$

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

$y_1 - y_2 = \Delta y = \frac{m_1 \lambda}{2} - \frac{m_2 \lambda}{2}$

=> $y_1 - y_2 = \Delta y = \frac{16 *480*10^{-9}}{2} - \frac{6 *480*10^{-9}}{2}$

=> $y_1 - y_2 = \Delta y = 5 (480*10^{-9})$

=> $\Delta y = 2.4 *10^{-6} \ m$

8 0

3 years ago

Give 1 real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is c

OverLord2011 [107]

Answer:

On real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is constant is when making a serve with a lawn tennis racket

How It is an example of impulse is that when a serve is made by moving the bat slowly, the lawn tennis player uses less force and the ball is in contact with the string for longer a period

When however, the lawn tennis player moves the racket faster, with the strings of the racket highly tensioned he uses more force and the ball also spends less time on the racket to produce the same momentum

Explanation:

The impulse of a force, ΔP is given by the following formula;

ΔP = F × Δt

Where ΔP is constant, we have;

F ∝ 1/Δt

Therefore, for the same impulse, when the force is increased, the time of contact is decreases and vice versa.

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3 years ago

Evidence on why we don’t hear solar eruptions on Earth.

sukhopar [10]

Because sound waves don't travel through the vaccume of space. Hope this helped

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2 years ago

Read 2 more answers

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI

OverLord2011 [107]

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

<u>Explanation: </u>

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } \times \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} \times 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^{-2}$

So, the work done can be expressed in $k g m s^{-2}$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2}$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^{2} / \mathrm{s}^{2}$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2} \times \frac{1000 \mathrm{g}}{1 \mathrm{kg}} \times \frac{100 * 100 \mathrm{cm}^{2}}{m^{2}}=3200 \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2}$

Thus, work done in ergs is 3200 ergs.

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3 years ago

State characteristics of the images formed by pin - hole camera​

State characteristics of the images formed by pin - hole camera