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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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ohaa [14]

Rocks within Earth both expand and contract as P waves pass

Explanation:

Rocks within the earth both expands and contracts as P-waves passes through them. P-waves are elastic waves.

  • Elastic waves behaves in such a way that they do not cause permanent deformation of rocks.
  • They can be said to cause elastic deformation when they travel through rocks.
  • They simply temporarily expand and contract the rock within a short period by causing the vibration of particles of the medium.
  • After a short while, the rock returns back to its original position as if nothing has happened to it.
  • These elastic waves are better called seismic waves.
  • P-waves are primary waves that can travel through any medium.

Learn more:

Earthquakes brainly.com/question/11292835

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6 0
3 years ago
: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma
e-lub [12.9K]

Answer:

x_1 = 3.74m

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

a_s = \dfrac{F}{m}

a_s = \dfrac{8.2}{12}

a_s = 0.68\ m/s^2

F = m a_m

a_m = \dfrac{F}{m}

a_m = \dfrac{8.2}{70}

a_m = 0.12\ m/s^2

x_1+x_2 = 25

\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25

(a_c+a_m)t^2=50

(0.12+0.68)t^2=50

t = \sqrt{\dfrac{50}{0.8}}

t = 7.90 s

x_1 = \dfrac{1}{2}a_ct^2

x_1 = 0.5\times 0.12 \times 7.90^2

x_1 = 3.74m

5 0
3 years ago
Which two of the following involve the same energy transfer. Assume that the same substance and the same mass is involved in all
Elanso [62]
B. evaporation
c. condensation

They are opposite processes that involve the same transfer of energy
3 0
3 years ago
Question 6. B) and c)
maks197457 [2]
The formula for both is v(t) = v0 + a*t

b) v(8) = 0 + 6m/s^2 *8s = 48 m/s

now we know the beginning (2) and end speed (14), but not the time:

c) 14 = 2 + 1.5*t => t = (14-2)/1.5 = 8 seconds
4 0
4 years ago
The electric potential at the surface of a charged conductor _______.
frez [133]

Answer:

The electric potential at the surface of a charged conductor<u> is always such that the potential is zero at all points inside the conductor.</u>

Explanation:

Each point on the surface of a balanced charged conductor has the same electrical potential.

The surface on any charged conductor in electrostatic equilibrium is an equipotential surface. Since the electric field is equal to zero inside the conductor, the electric potential at any point inside and on the surface is equivalent to its value.

6 0
3 years ago
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