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astraxan [27]
4 years ago
11

C. 400 km/min

Physics
1 answer:
Nataly_w [17]4 years ago
5 0
A is the answer i hope
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Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll
Andrew [12]

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

Q = \dfrac{4mME }{(m+M)^2}

However; from the total stopping power & power loss of the electron;

\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss  =\dfrac{82 \times 1.9}{800}

= 0.19475

b)

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

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4 0
3 years ago
The force between a pair of .005 charges is 750 N. What is the distance between them?
Ganezh [65]

Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?

Answer:

17.32 m

Explanation:

From coulomb's Law,

F = kqq'/r²........................... Equation 1

Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.

make r the subject of the equation above

r = √(kqq'/F)..................... Equation 2

From the question,

Given: q = q' = 0.005 C, F = 750 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(9.0×10⁹×0.005×0.005/750)

r = √(300)

r = 17.32 m.

Hence the distance between the pair of charges = 17.32 m

6 0
3 years ago
if a current of 0.3 ampere flows through a conductor of resistance of 11.7 ohm's,the voltage across the ends of the conductor is
Gnesinka [82]

Here,

Resistance = 11.7

Ampere = 0.3

Voltage = ?

Now,

R = V/A

11.7 = V/0.3

11.7*0.3 =V

V = 3.51

I hope it's help you...

Mark me as brainliest...

7 0
3 years ago
Answer these questions plz​
lozanna [386]

1. The property of a conductor by virtue of which it posses the flow of electric current through it is called resistance.

2. The resistance of a conductor depends on the cross sectional area of the conductor and it's resistivity.

3.This id due to the fact that the resistance of a wire is inversely proportional to the square of its diameter.

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Marine magnetic anomalies result from seafloor spreading in conjunction with
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Earths magnetic field
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