Charges of 7.2nC and 6.7nC are 32 cm apart. Find the equilibrium position for a -3.0nC charge.

The force of attraction between any two objects is known as

Leona [35]

Answer:

Gravity :)

Explanation:

7 0

3 years ago

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A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine

Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = $\frac{4}{3}\pi(\frac{d}{2})^3$

or

Volume of the balloon = $\frac{4}{3}\pi(\frac{0.296}{2})^3$

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = $\frac{3}{2}\times K_BT$

where,

Boltzmann constant, $K_B=1.3807\times10^{-23}J/K$

Average kinetic energy = $\frac{3}{2}\times1.3807\times10^{-23}\times292$

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = $\frac{3RT}{m}$

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = $\frac{3\times8.314\times292}{0.004}$

or

rms speed = 1349.35 m/s

5 0

3 years ago

Which color of the visible light waves has the highest energy level?

sveticcg [70]

Red..................

3 0

2 years ago

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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to

nordsb [41]

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

$5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s$

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

4 0

3 years ago