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seropon [69]
3 years ago
6

In the diagram, q1 = +6.39*10^_9 C and

Physics
1 answer:
ivanzaharov [21]3 years ago
4 0

Answer:

The electric potential will be "259.695 volt".

Explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

q_1=6.39\times 10^{-9} \ C

q_2=3.22\times 10^{-9} \ C

AP=(0.150+0.250)

      =0.40 \ m

BP=0.25 \ m

Now,

At point P, the electric potential will be:

⇒ V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}

By putting values, we get

⇒     =9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]

⇒     =259.695 \ Volt

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Hatshy [7]
The last one is correct (D)
8 0
3 years ago
In a liquid with a density of 1400 kg/m3 , longitudinal waves with a frequency of 370 Hz are found to have a wavelength of 8.40
VARVARA [1.3K]

Answer:

Bulk modulus = 1.35 × 10^{10} Pa

Explanation:

given data

density = 1400 kg/m³

frequency = 370 Hz

wavelength = 8.40 m

solution

we get here bulk modulus of the liquid that is

we know Bulk Modulus = v^2*\rho   ...............

here \rho is density i.e 1400 kg/m³

and v is =  frequency × wavelength

v = 370 × 8.40 = 3108 m/s

so here bulk modulus will be as

Bulk modulus = 3108² × 1400

Bulk modulus = 1.35 × 10^{10} Pa

3 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
Sergeeva-Olga [200]

Answer:

33.23 m

Explanation:

At the point where both objects will meet, the vertical height will be equal.

From the equations of motion, the vertical height of the body falling at any time is given as

(y - y₀) = ut + ½gt²

y = vertical height at any time T

y₀ = initial height of the object = 81.5 m

u = initial velocity = 0 m/s (body was dropped)

g = -9.8 m/s²

(y - 81.5) = 0 - 4.9T²

y = 81.5 - 4.9T² (eqn 1)

For an object thrown up, the vertical height of the body at any time, t, is given as

(y - y₀) = ut + ½gt²

y = vertical height of the object at any time t

y₀ = initial height of the object = 0 m

u = initial velocity = 40 m/s

g = -9.8 m/s²

y = 40t - 4.9t² (eqn 2)

At the point where the two objects meet, we equate eqn 1 and eqn 2

y = y

81.5 - 4.9T² = 40t - 4.9t²

But T = (t + 2.2) (Since object 2 was dropped 2.2 s after object 1)

81.5 - 4.9(t + 2.2)² = 40t - 4.9t²

81.5 - 4.9(t² + 4.4t + 4.84) = 40t - 4.9t²

81.5 - 4.9t² - 21.56t - 23.716 = 40t - 4.9t²

81.5 - 21.56t - 23.716 - 40t = 0

57.784 = 61.56t

t = (57.784/61.56) = 0.93866 = 0.94 s

Therefore, the vertical height at t = 0.93866 s is

y = (40×0.93866) - 4.9(0.93866²) = 33.23 m

Hope this Helps!!!

5 0
3 years ago
A complex sound’s quality depends on the mix of harmonics or overtones added to the ______________ frequency
Brilliant_brown [7]
Fundamental frequency is the "base frequency" upon which the sound is built.
4 0
3 years ago
Just think about this.
diamong [38]
This ain’t the place, bud. If you have a QUESTION, then you can post it here.
8 0
3 years ago
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