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AleksandrR [38]
3 years ago
11

If a certain gas occupies a volume of 12 l when the applied pressure is 6.0 atm , find the pressure when the gas occupies a volu

me of 3.0 l .
Physics
1 answer:
dolphi86 [110]3 years ago
5 0
From Boyle's law, the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature. 
Therefore; P1V1 =P2V2; where PV is a constant
hence; 12 × 6 = 3× p2
           p2 = 72/3
                = 24 atm
Therefore; the new pressure will be 24 atm
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The distance between two successive peaks on adjacent waves is its
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Answer:

Wavelength

Explanation:

Wavelength is the distance between two corresponding consecutive phases of a waveform. It is usually represented by λ in the mathematical expressions.

A continuous propagating wave repeats its wavelength over the distance.

A wave has crest and trough with respect to time and space.

Wave is defined as a disturbance of any parameter repeated in a cyclic manner over the given time.

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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
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Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

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2 years ago
A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocit
Dmitry [639]

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

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F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

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F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

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