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777dan777 [17]
3 years ago
13

Rami uses a disposable camera that has a flash. When he wants to take a picture, he holds a button and hears a rising whining so

und. When a small light glows, the flash is ready. He presses the shutter button, the flash lights up, and he takes his picture.
Which statement best describes how a capacitor is useful in the camera?

It protects the circuits from a burst of energy by storing charge.
It protects the circuits from a steady current by storing charge.
It stores energy and delivers it in a short burst.
It stores energy and delivers it in a steady current.
Physics
1 answer:
ivann1987 [24]3 years ago
6 0
<span>It stores energy and delivers it in a short burst.

The whirring sound is produced by the charging of the capacitor. A capacitor is an electrical component which is capable of storing charge. When the capacitor stores charge, it is storing energy. After doing so, the capacitor releases the electrical energy that it had stored as light energy, which is seen as the flash of the camera. It must do so in a burst, because the intensity of the flash is very high and would require a high amount of energy to maintain.
</span>
You might be interested in
Pls look at the photo above, thanks
frosja888 [35]
V=s/t. All you need to do is
s=v×t=330×1.8
4 0
3 years ago
Three children are riding on the edge of a merry-go-round that is a solid disk with a mass of 102 kg and a radius of 1.53 m. The
Mnenie [13.5K]

Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

They are all initially riding at the edge of the merry go round

Then, R1 = R2 = R3 = R = 1.7m

Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

Angular velocity of Merry go round

ωi = 22 rpm

If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

Using conservation of angular momentum

Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round  Then,

L(initial) = L(final)

Ii•ωi = If•ωf

So we need to find the initial and final moment of inertia

NOTE: merry go round is treated as a solid disk then I= ½MR²

I(initial)=½MR²+M1•R²+M2•R²+M3•R²

I(initial) = ½MR² + R²(M1 + M2 + M3)

I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

I(initial) = 151.725 + 1.7²(83)

I(initial) = 391.595 kgm²

Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

I(final) = 151.725 + 158.95

I(final) = 310.675 kgm²

Now, applying the conservation of angular momentum

L(initial) = L(final)

Ii•ωi = If•ωf

391.595 × 22 = 310.675 × ωf

Then,

ωf = 391.595 × 22 / 310.675

ωf = 27.73 rpm

Answer: So, the final angular momentum is 27.73 revolution per minute

7 0
3 years ago
The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi
vlabodo [156]

Answer:

Smallest drop: Water

Largest drop: Dirt

Explanation:

The heat needed to change the temperature of a sample is:

Q=cm\Delta T (1)

with Q the heat (added(+) or removed(-)), c specific heat, m the mass and \Delta T the change in temperature of the sample. So, if we solve (1) for

Sample A:

\Delta T=-\frac{Q}{cm} =\frac{Q}{4186*4.0}

\Delta T=-\frac{Q}{16744}

Sample B:

\Delta T=-\frac{Q}{cm} =\frac{Q}{2700*2.0}

\Delta T=-\frac{Q}{5400}

Sample C:

\Delta T=-\frac{Q}{cm} =\frac{Q}{1050*9.0}

\Delta T=-\frac{Q}{9450}

Note that the numbers 16744, 5400, 9450 are in the denominator of the expression -\frac{Q}{cm} that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.

So, the smallest drop is Sample A and the largest is Sample C.

(Important: The minus sign of \Delta T implies the temperature is dropping)

8 0
3 years ago
Suppose that a ball is released from the window of a train that is moving with constant velocity.  The path of the ball, as obse
barxatty [35]

True, the path of the ball, as observed from the train window, will be a horizontal straight line.

An object projected from a certain height has a parabolic path when observed from a fixed point.

However, if the reference point is moving at the same velocity as the object, the path of the object's motion appears to be a straight line.

When the ball is released from the window of the train, it will move at the same constant velocity as the train, and the path of the ball's motion observed from the train window will be a straight line.

Thus, we can conclude that the given statement is true. The path of the ball, as observed from the train window, will be a horizontal straight line.

Learn more about path of motion of objects here: brainly.com/question/82610

7 0
2 years ago
A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and c
tankabanditka [31]

Answer:

The magnetic field strength inside the solenoid is 5.026\times10^{-3}\ T.

Explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

B=\dfrac{\mu_{0}NI}{l}

Where, N = Number of turns

I = current

l = length

Put the value into the formula

B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}

B=0.005026=5.026\times10^{-3}\ T

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is 5.026\times10^{-3}\ T.

4 0
3 years ago
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