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3 years ago

Which equation best describes the law of conservation of momentum?

2 answers:

8 0

The law of conservation of momentum states that for two objects colliding in an isolated system, the total momentum before and after the collision is equal. Momentum should be conserved. Hope this answers the question. Have a nice day.

allochka39001 [22]3 years ago

6 0

Answer:

pi = pf

Explanation:

its the equation

You might be interested in

What’s 3 times 10 to the 8th power divided by 2.45 times 10 to the 9th power

Virty [35]

0.12244898 is the value obtained when solving the given.

Explanation:

Given:

3 times 10 to the 8th power can be expressed in equation format as $3 \times 10^{8}$ (3 times symbolizes ‘multiplication’ then it is to the tenth power of 8)

2.45 times 10 to the 9th power can be expressed in equation format as $2.45 \times 10^{9}$ (2.45 times symbolizes ‘multiplication’, then it is to the tenth power of 9)

Asked to find the solution when dividing the above,

$\frac{3 \times 10^{8}}{2.45 \times 10^{9}}=\frac{3}{2.45} \times 10^{8-9}$

When the tenth power of any value goes from denominator to numerator and vice-versa, presents in opposite sign to that it possess (like when ‘ $10^{9}$ ' goes to numerator changed as ' $10^{-9}$ ').

Now by solving the above equation, we get

$\frac{3}{2.45} \times 10^{8-9}=1.22 \times 10^{-1}=\frac{1.22}{10}=0.12244898$

5 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:

Given:

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,

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

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

4 0

3 years ago

A negative charge of -2.0 C and a positive charge of 3.0 C are separated by 80 m. What is the electrostatic force between the tw

faltersainse [42]

Answer:

1. 8437500 N

2. The force between the two charges is attractive.

Explanation:

1. Determination of the force between the two charges.

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

Distance apart (r) = 80 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 2 × 3 / 80²

F = 5.4×10¹⁰ / 6400

F = 8437500 N

Thus, the force of attraction between the two charges is 8437500 N

2. From the question given, the charges are:

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.

Thus the force between them is attractive.

6 0

3 years ago

What element is produced when the isotope bismuth-213 emits a beta particle

rusak2 [61]

Answer:

Explanation:

4 0

3 years ago

What is the density of an object with a mass of 60g and a volume of 2cm3?

zaharov [31]

m= 60g = 60/1000 Kg = 0.06 Kg v = 2cm3 = 2 * (0.01^3) m3 = 2 *10^-6 m3 Density= m/v = 6 * 10^-2 / 2 *10^-6 = 3 *10^4 Kg/m3

8 0

3 years ago