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andrey2020 [161]
3 years ago
13

Which step in an investigation of arson helps insurance claims the most?

Physics
1 answer:
mixer [17]3 years ago
4 0

The step in an investigation of arson that help the insurance claims the most is the samples of ignition sources or the processing of evidence at the scene. The people with the insurance claims are careful in gathering investigation data for they will used it as a basis for their claim.

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If earth stops rotating then what will be the value of 'g' at poles
Marianna [84]
The value of 'g' is not affected by rotation at any place on Earth.
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3 years ago
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
bija089 [108]

This question is incomplete, the complete question is;

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.

At what rate is the magnetic field changing?

Answer:

the magnetic field changing at the rate of 9.33 m T/s

Explanation:

Given the data in the question;

Electric field E = 7 mV/m

radius r = 1.5 m

Now, from Faraday law of induction;

∫E.dl = d∅/dt

E∫dl = A( dB/dt )

E( 2πr ) = πr² ( dB/dt )

( 0.007 ) = (r/2) ( dB/dt )

( 0.007 ) = 0.75 ( dB/dt )

dB/dt = 0.007 / 0.75

dB/dt = 0.00933 T/s

dB/dt = ( 0.00933 × 1000) m T/s

dB/dt = 9.33 m T/s

Therefore, the magnetic field changing at the rate of 9.33 m T/s

5 0
3 years ago
Amount of pressure of liquid increases with ?​
Pachacha [2.7K]

Answer: Pressure increases as the depth increases.

3 0
2 years ago
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A chair of weight 90.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F =
Lelu [443]
I post an image instead

3 0
2 years ago
A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh
Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = \frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0
2 years ago
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