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klasskru [66]
3 years ago
10

Which of the following statements accurately describe a scientific law?

Physics
1 answer:
Rufina [12.5K]3 years ago
3 0

Answer:

The first statement appears most accurate.

The last statement also appears accurate, but it does not preclude any contradictory evidence.

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During normal beating, the heart creates a maximum 3.60 mV potential across 0.250 m of a person's chest, creating a 1.00 Hz elec
grandymaker [24]

Answer:

The maximum electric field strength is 0.0144 V/m.

Explanation:

Given that,

Electric potential created in the heart, V = 3.6 mV

Distance, d= 0.25 m

Frequency of the the electromagnetic wave, f = 1 Hz

We need to find the maximum electric field strength created. We know that the electric potential is given by :

V=Ed

E is the maximum electric field strength

E=\dfrac{V}{d}\\\\E=\dfrac{3.6\times 10^{-3}\ V}{0.25\ m}\\\\E=0.0144\ V/m

So, the maximum electric field strength is 0.0144 V/m. Hence, this is the required solution.

5 0
3 years ago
How is an experiment is actually conducted
NeX [460]
By using the scientific method
3 0
3 years ago
Tick (3) the correct statement about electrostatic charges.
Radda [10]

Answer:

similar type of electric charges attract one another

I think this is a coorect staement

6 0
3 years ago
Read 2 more answers
The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbe
8090 [49]

Answer:

579600J

Explanation:

Given parameters:

Height of the building  = 828m

Weight of the man  = 700N

Unknown:

Work done by the man  = ?

Solution:

The work done by the man is the same as the potential energy expended.

Work done:

            Work done  = Weight x height  = 700 x 828

       Work done  = 579600J

7 0
3 years ago
Yellow light of wavelength 590 nm passes through a diffraction grating and makes an interference pattern on a screen 80 cm away.
riadik2000 [5.3K]

Answer:B

Explanation:

Given

Wavelength of light \lambda =590\ nm

Screen distance L=80\ cm

First fringe is at a distance y_1=1.9\ cm

No of lines per mm is given by N

N=\frac{1}{d}

where d=slit width

From N-slits Experiment

\sin \theta _m=\frac{m\lambda }{d}

d=\frac{m\lambda }{\sin \theta _m}-----1

Position of bright fringe is given by

y=\tan \theta _m\cdot L

\tan \theta _m=\frac{y}{L}

\theta _m=\tan^{-1}(\frac{y}{L})

Put the value of \theta _m  in eq. 1

d=\frac{m\lambda }{\sin (\tan^{-1}(\frac{y}{L}))}

Therefore N=d^{-1}

N=\frac{\sin (\tan^{-1}(\frac{y}{L}))}{m\lambda }

for m=1

N=\frac{\sin (\tan^{-1}(\frac{1.9\times 10^{-2}}{0.8}))}{1\times 590\times 10^{-9}}

N=40243\ line/m

N=40\ line/mm

   

4 0
4 years ago
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