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natita [175]
3 years ago
13

A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc

uses on the same point, but changes the speed to 1/250 s. What needs to happen to the diameter of the aperture, if it is still properly exposed?a. Nothing b. D' = 16 mm c. D'=1.0 mm d. D' = 8.0 mm e. D'=2.0 mm
Physics
1 answer:
hoa [83]3 years ago
4 0

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

E V = log_2(\dfrac{N^2}{t})

N is the diameter of the aperture and t is the time of exposure

now,

log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})

\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}

inserting all the values

\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}

      N₂² = 4

      N₂ = 2 mm

hence , the correct answer is option E

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si tienes 2 circuitos uno en serie y otro en paralelo con el mismo valor en sus resistencias, para cuál circuito la resistencia
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Answer:

circuito paralelo

Explanation:

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A disc on a frictionless axle, starting from rest (0 rpm) can spin up to a rotation rate of 3820 rpm in a period of 2 seconds. (
Eddi Din [679]

Answer:

1000 Nm

2000 Nm

1.00007 seconds

Explanation:

I = Moment of inertia = 5 kgm²

\alpha = Angular acceleration

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\omega_i = Initial angular velocity

t = Time taken

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=5\times 200\\\Rightarrow \tau=1000\ Nm

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If \alpha=400\ rad/s^2

\tau=I\alpha\\\Rightarrow \tau=5\times 400\\\Rightarrow \tau=2000\ Nm

The torque of the disc would be 2000 Nm

From equation of rotational motion

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It would take 1.00007 seconds to reach 3820 rpm

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4 years ago
TRUE OR FALSE<br><br><br> An object will slow down and stop as long as no forces act upon it
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Answer:

False

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Read 2 more answers
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