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cricket20 [7]
2 years ago
12

A teacher wants to demonstrate that the radioactive source emits alpha beta and gamma radiation.

Physics
1 answer:
Mashcka [7]2 years ago
6 0

It is possible to distinguish between these types of radiation by the use of an electrostatic field.

<h3>What is a radioactive source?</h3>

A radioactive source is a source that emits radiation such as alpha, beta and gamma radiation. We can be able to distinguish between these types of radiation by the use of an electrostatic field.

The gamma rays is undeflated by the field, the alpha ray is deflated to the negative part of the field while the beta rays is deflated by the positive part of the field.

Learn more about radioactive source: brainly.com/question/12741761?

#SPJ1

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Elements in the periodic table are arranged in order of increasing atomic number

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Which answer choice provides the best set of labels for Wave A and Wave B?
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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
vladimir1956 [14]

Answer:

Explanation:

kinetic energy required = 1.80 MeV

= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 2.88 x 10⁻¹³ J

If v be the velocity of proton

1/2 x mass of proton x v² = 2.88 x 10⁻¹³

= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³

v² = 3.45 x 10¹⁴

v = 1.86 x 10⁷ m /s

If V be the potential difference required

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V = 1.8 x 10⁶ volt .

3 0
3 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

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3 years ago
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