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elena-14-01-66 [18.8K]
4 years ago
6

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics
2 answers:
Fynjy0 [20]4 years ago
8 0
D.
The reading between 7N and 8N would have to be 7.5N. Answers A and B are much to small and answer C is way to big.
vesna_86 [32]4 years ago
7 0

A, B, or C are not between 7 and 8.  So it must be D .

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Nadia drove home while singing with a recording of her favorite songs.This is an example of?(psychology)
lianna [129]

Answer:

Music

Explanation:

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The annoying sound from a mosquito is produced when it beats its wings at the average rate of 600. wingbeats per second. What is
jok3333 [9.3K]
The frequency is exactly the rate at which the bug beats its wings.
If that "600" that you mentioned is 600 beats per second, then
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Drag the tiles to the boxes to form correct pairs. The table shows the distances between a star and three celestial objects. Use
aliina [53]

Answer:

Explanations :

4 1
3 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

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7 0
3 years ago
Work and Energy
Sergeeva-Olga [200]

a) W_1 = 2332 J, W_2= 2332 J

The work done by the student in each trial is equal to the gravitational potential energy gained by the student:

W=mg\Delta h

where

m = 68 kg is the mass of the student

g = 9.8 m/s^2 is the acceleration of gravity

\Delta h is the gain in height of the student

For the first student, \Delta h = 3.5 m, so the work done is

W_1 = (68)(9.8)(3.5)=2332 J

The second student runs up to the same height (3.5 m), so the work done by the second student is the same:

W_2 = (68)(9.8)(3.5)=2332 J

2) P_1 = 204.6 W, P_2 = 274.4 W

The power exerted by each student is given by

P=\frac{W}{t}

where

W is the work done

t is the time taken

For the first student, W_1 = 2332 J and t=11.4 s, so the power exerted is

P_1 = \frac{W_1}{t_1}=\frac{2332 J}{11.4 s}=204.6 W

For the second student, W_2 = 2332 J and t=8.5 s, so the power exerted is

P_2 = \frac{W_2}{t_2}=\frac{2332 J}{8.5 s}=274.4 W

5 0
3 years ago
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