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4 years ago

How does the total mass before a nuclear reaction compare with the total mass after a nuclear reaction?

1 answer:

6 0

Atomically speaking, there is a small mass defect between the parent molecule and the one “achieved” after the reaction. This small amount lost is actually the energy converted and from the matter through the reaction and released as a by-product; when observed on the macro scale, this small amount of material lost results in the large atomic explosions and radioactive fallout that occurs following a nuclear detonation.

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Light passes from a material A, which has an index of refraction of 4/3, into material B, which has an index of refraction of 5/

katrin2010 [14]

Answer:

(b) 32.2°

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 30.0° )

${\theta_r}$ is the angle of refraction ( ? )

${n_r}$ is the refractive index of the refraction medium (Material B, n=5 / 4)

${n_i}$ is the refractive index of the incidence medium (Material A, n=4 / 3)

Hence,

${\frac {4}{3}}\times {sin30.0^0}={\frac {5}{4}}\times{sin\theta_r}$

Angle of refraction = $sin^{-1}0.5333$ = 32.2°

5 0

4 years ago

You’re still in a redecorating mood and want to move a 35kg bookcase to a different place in the living room. You exert 58 N on

CaHeK987 [17]

Answer: 0.16

Explanation:

Newton's second law states that the resultant of the forces exerted on the bookcase is equal to the product between the mass (m) and the acceleration (a):

$F-F_f = ma$

where

F = 58 N is the force applied

Ff is the frictional force

Substituting, we find the frictional force

$F_f = F-ma=58 N-(35 kg)(0.12 m/s^2)=53.8 N$

The frictional force has the form:

$F_f = \mu mg$

where $\mu$ is the coefficient of kinetic friction. Re-arranging the formula, we can find the coefficient:

$\mu=\frac{F_f}{mg}=\frac{53.8 N}{(35 kg)(9.8 m/s^2)}=0.16$

4 0

4 years ago

If you could live any where except for where you are now where would you live

laila [671]

Answer:

I would live in the Atlantic ocean on a lux liner

Explanation:

4 0

3 years ago

Read 2 more answers

in terms of the wavelength of the sound wave, about how far below the open end of the resonance tube is the first resonant posit

notka56 [123]

The first resonant position below the open end of the resonance tube is; <em><u>one-quarter of the wavelength</u></em>

In the event of the first resonant position in a resonance tube, there will be a maximum air displacement which is only one antinode right at the open end where the motion is constrained.

However, there will be no displacement at the closed end which means another one node right at the closed end where air is halted.

This means that the standing wave will have one-quarter of the wavelength in the test tube.

Thus;

L = ¼λ

Read more at; brainly.com/question/17086525

8 0

2 years ago

A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri

vekshin1

Electric charge on the plastic cube: $1.3\cdot 10^{-7}C$

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

$V(r)=\frac{kQ}{r}$

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

$Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V$

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

$qV' = \frac{1}{2}mv^2$