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dalvyx [7]
3 years ago
11

100 POINTS UP FOR GRABS FOR ANYONE WILLING TO HELP! PLEASE!

Physics
1 answer:
Aloiza [94]3 years ago
5 0
I can’t read what’s written on the documents in the pictures
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What is the wavelength of light (nm) that has a frequency of 6.44 x 1013 s-1?
lesantik [10]
4660

Hope this helped!
STSN
4 0
3 years ago
Read 2 more answers
The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec
marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

\epsilon = \frac{\sigma}{\sqrt{n} }

To reduce the standard error to 0.03 s, let the additional number of trials = x

0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials

Therefore, the additional trials needed is 48 trials.

6 0
3 years ago
how much work does tension do on the laptop as it moves 2.0 mm ? express your answer with the appropriate units.
8090 [49]

The work done by tension force of 14N applied on the laptop by a rope as it moves 2.0 mm up the slope is 0.028 J

W = F d cos θ

W = Work done

F = Force

d = Displacement

θ = Angle between force and displacement vector

F = 14 N

d = 2 mm = 0.002 m

θ = 0

W = 14 * 0.002 * 1

W = 0.028 J

Work done is the change in energy of an object. So if an object moves a certain distance, work is done on the object. If the force and displacement are perpendicular to each other there is no work done on the object.

Therefore, the work done by tension on the laptop is 0.028 J

To know more about work done

brainly.com/question/12834956

#SPJ4

8 0
1 year ago
How many significant figures does the value 0.080 have? 1 2 3 4
lara [203]
2. The zeros in front do not matter
3 0
4 years ago
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
hoa [83]

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

4 0
3 years ago
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