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3 years ago

In the nature-nurture debate, nature refers to the influence of the environment.

1 answer:

7 0

I believe that statement is wrong because nature is when you're born a certain way rather than being influenced by outside forces like you environment which would be nurture.

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Three equal 1.60-μCμC point charges are placed at the corners of an equilateral triangle with sides 0.800 mm long. What is the p

Pepsi [2]

Answer:

= 0.086J

Explanation:

we are given three equal charges q₁ q₂ q₃ = 1.60 μC

sides of triangle = 0.800m

U = 1/4πε₀(3q²/r)

1/4πε₀ = 9.0 × 10⁹ N.m²/C²

U = 9.0 × 10⁹ ((3 * 1.60 × 10⁻⁶)² / (0.800))

= 0.086J

3 0

3 years ago

What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movem

Step2247 [10]

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

= 1/5 ∫ 3*10^-3 d(t) + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

The reading of the voltmeter ( V )

V = IR

= 0.96 mA * 21 k Ω = 20.16 v

3 0

2 years ago

Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a

BartSMP [9]

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1 ....................1

2 × t × μ = (m2+0.5) × λ2 ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

$\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}$ ..............3

put here value and we get

$\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}$

$\frac{m1+0.5}{m2+0.5}$ = 1.4

$\frac{m1+0.5}{m2+0.5} = \frac{7}{5}$ ...................4

so we here from equation 4

m1+0.5 = 7

m1 = 3 .................5

m2+0.5 = 4

m2 = 2 .................6

so now put value in equation 1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) × 477.1

solve it we get

thickness t = 528.433 nm

4 0

3 years ago

A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a

Colt1911 [192]

Answer:

The intensity of sound wave at the surface of the sphere $I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

Explanation:

B = Bulk modulus

Intensity, $I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }$

The amplitude of oscillation of the sphere is given by:

$P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\$

$A = \triangle R\\$

Substitute v and A into Pmax

$P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}$

$I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }$

$P_{total} = 4\pi R^{2} I$

$P_{total} =4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }$

The intensity of the sound wave at a distance is given by:

$I = \frac{P_{total} }{4\pi d^{2} }$

$I = 4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

5 0

3 years ago

A ball is thrown vertically upward from the edge of a bridge 22.0 m high with an initial speed of 16.0 m/s. The ball falls all t

KonstantinChe [14]

To solve this problem we will apply the concepts related to the kinematic equations of motion. We will start calculating the maximum height with the given speed, and once the total height of fall is obtained, we will proceed to calculate with the same formula and the new height, the speed of fall.

The expression to find the change in velocity and the height is,

$v_f^2-v_0^2 = -2gh$