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2 years ago

How is radioactivity produced

Physics

1 answer:

aev [14]2 years ago

8 0

Explanation:

It is a physical observed phenomenon, where an atom is emitting type of particles such as, alpha or beta followed by gamma radiations, which change the radioactive atom.

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Increasing temperature increases solubility so that 480 grams (g) of sugar makes a saturated solution in 100 milliliters (mL) of

timofeeve [1]

150 ml of saturated solution

5 0

2 years ago

Read 2 more answers

A bullet with a mass ????b=13.5mb=13.5 g is fired into a block of wood at velocity ????b=253vb=253 m/s. The block is attached to

Lorico [155]

Answer:

450 grams

Explanation:

Given:

mass of the bullet, m = 13.5 g = 0.0135 kg

velocity of the bullet, v = 253 m/s

spring constant of the spring, k = 205 N/m

Compression of the spring, x = 35.0 cm = 0.35 m

Now, the kinetic energy of the moving system (bullet + board) will tend to move the spring

thus,

let velocity of the system, V

now, applying the concept of conservation of momentum, we have

mv = (M + m)V

where,

M is the mass of the block

thus,

V = mv/(M + m)

now,

the kinetic energy of the system = (1/2)(M + m)V²

the kinetic energy of the system = (1/2)(M + m)(mv/(m + M))²

Energy gained by the spring = (1/2)kx²

now,

equating both the energies, we get

(1/2)(M + m)(mv/(m + M))² = (1/2)kx²

(mv)²/(m + M) = kx²

on substituting the values, we get

(0.0135 × 253)²/(0.0135 + M) = 205 × (0.35)²

11.66/(0.0135 + M) = 25.1125

M = 0.450 kg = 450 grams

8 0

3 years ago

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$