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Sladkaya [172]
3 years ago
7

Most photographs will be made using a shutter speed of 1/60 or faster.

Physics
2 answers:
AysviL [449]3 years ago
7 0

Answer:

True

Explanation:

The shutter speed has a long way in determining the quality of image captured. For a high quality image, the lens requires a low amount of light stimulating it. Cameras with slow shutter speed allow more light to get to the lens of the camera thereby producing a blurred image. 1/60 is a tolerable shutter speed if you desire a sharp image.

Have a great day ahead

kari74 [83]3 years ago
5 0
True

Most photographs will be made using a shutter speed of 1/60 or faster.
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One part of a freely swinging magnet always points to the Earth's magnetic pole in the Northern Hemisphere.
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An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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7 0
3 years ago
How much force is required to accelerate a 12 kg mass at 5 m/s 2
Savatey [412]

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

4 0
2 years ago
A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr
PSYCHO15rus [73]

Answer:

ω = 2.1 rad/sec

Explanation:

  • As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
  • This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
  • Now, we need to ask ourselves: what supplies this force?
  • In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
  • We know that this force can be expressed as follows:

       f_{frs} = \mu_{s} * F_{n} (1)

      where μs = coefficient of static friction between the rock and the merry-

      go-round surface = 0.7, and Fn = normal force.

  • In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
  • Fn = m*g (2)
  • This static friction force is just the same as the centripetal force.
  • The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

       F_{c} = m* \omega^{2}*r (3)

  • Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

       \omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec

  • This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.
4 0
3 years ago
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