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3 years ago

Which career is most concerned with the study of radioactive isotopes?

1 answer:

3 0

<span>The career that is most concerned with the study of radioactive isotopes is chemistry. You need to have passed AP chemistry to actually deal with the isotopes. In a quick reference, isotopes are when the atom has difference amount of neutrons, making the atomic mass differ.</span>

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Which type of electromagnetic radiation may damage central nervous system

Furkat [3]

Answer:

microwaves

Explanation:

microwaves do emit radiation, technically speaking, but it's not the DNA-damaging radiation we're used to hearing about. Microwaves, along with radio waves from (you guessed it) radio and cell phone towers, are types of non-ionizing radiation.

5 0

3 years ago

Read 2 more answers

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

7 0

3 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

6 0

3 years ago

Read 2 more answers

Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is r

Naddik [55]

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between both objects.

D is now reduced by a factor of 5, meaning Dnew = D/5 we get

Fgravitynew = G*(mass1*mass2)/(D/5)² =

= G*(mass1*mass2)/(D²/25) =

= 25* G*(mass1*mass2)/D² = 25* Fgravity

the new force of gravity/attraction is 25×16 = 400 units.

4 0

2 years ago

What is the energy in space left over from the big bang called

Lilit [14]

It is called the CMBR, which stands for cosmic microwave background radiation. It was discovered by Arno Penzias and Robert Wilson in 1964.

5 0

3 years ago