All categories

3 years ago

What advantage does a modern camera offer a pinhole camera

Physics

1 answer:

AleksAgata [21]3 years ago

4 0

Answer:

Done

Explanation:

The differences are significant. The main disadvantages of a pinhole camera compared to a modern camera are:

Proper exposure requires a trial and error approach.

Requires manual control of shutter speed (difficult to control)

Poor image quality (due to diffraction effects).

Long exposures (lead to blurry images if the subject moves).

Fixed aperture or f-stop.

Advantages include:

Large depth of field.

Inexpensive

Robust and nearly 100% reliable.

If used properly can create artistic effects.

You might be interested in

Determine the linear velocity of an object with an angular velocity of 5.9 radians per second at a distance of 12 centimeters fr

Andre45 [30]

The linear velocity of a rotating object is the product of the angular velocity and the radius of the circular motion. Angular velocity is the rate of the change of angular displacement of a body that is in a circular motion. It is a vector quantity so it consists of a magnitude and direction. From the problem, the angular velocity is 5.9 rad per second and the radius is given as 12 centimeters. We calculate as follows:

Linear velocity = angular velocity (radius)
Linear velocity = 5.9 (12 ) = 70.8 cm / s

The linear velocity of the body in motion is 70.8 centimeters per second or 0.708 meters per second.

7 0

3 years ago

A stunt woman attempts to swing from the roof of a m 24 high building to the bottom of an identical building using a m 24 rope,

MakcuM [25]

Answer:

Height from ground is 8 m where string will break

Explanation:

Let the string makes some angle with the vertical after some instant of time

So here we have

$T - mg cos\theta = \frac{mv^2}{R}$

$T = mg cos\theta + \frac{mv^2}{R}$

now by energy conservation we have

$\frac{1}{2}mv^2 = mg(R cos\theta)$

$mv^2 = 2mgR cos\theta$

$T = mg cos\theta + 2mgcos\theta$

$T = 3mg cos\theta$

For string break down we have

$T = 2mg = 3mgcos\theta$

$cos\theta = \frac{2}{3}$

Now height from the ground is given as

$h = R - Rcos\theta$

$h = 24 - 24(\frac{2}{3})$

$h = 8 cm$

7 0

3 years ago

A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta

serg [7]

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

$mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s$

So, the final velocity of the bullet is 9 m/s.

5 0

3 years ago

Ethan pushes a wooden box across a carpeted floor. Then he pushes the same box across a smooth marble floor. Why does Ethan find

Ksju [112]

Answer:

B.The box experiences less friction on the marble floor

Explanation:

8 0

3 years ago

Read 2 more answers

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago