The psychological experience of pitch is related to a sound wave’s:.

If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of a

Artist 52 [7]

Answer:

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Explanation:

The power dissipation by an electrical circuit is given by the following formula:

Power Dissipation = (Voltage)(Current)

P = VI

but, from Ohm's Law, we know that:

Voltage = (Current)(Resistance)

V = IR

Substituting this in formula of power:

P = (IR)(I)

P = I²R ---------------- equation 1

Now, if we double the current , then the power dissipated by that circuit will be:

P' = I'²R

where,

I' = 2 I

Therefore,

P' = (2 I)²R

P' = 4 I²R

using equation 1

<u>Therefore, the power dissipated by the circuit will becomes four times of its initial value.</u>

4 0

3 years ago

A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o

Ad libitum [116K]

Answer:

The thickness of the oil slick is $1.95\times10^{-7}\ m$

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

$2nt= m\lambda$

Where, n = Index of refraction

t = thickness

$\lambda$ = wavelength

Put the value into the formula

$2\times1.28 \times t=1\times\times500\times10^{-9}$

$t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }$

$t=1.95\times10^{-7}\ m$

Hence, The thickness of the oil slick is $1.95\times10^{-7}\ m$

4 0

3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude

matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, $v_t$ = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

$v_t$ = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, $v_t$ = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0

3 years ago

You are a geologist studying the electrical field on a given region. You find that the electric field is vertical and perpendicu

Xelga [282]

Answer:

charge Qint = 7.17 10⁻⁴ C

Explanation:

For this problem we must use Gauss's law

F = ∫ E. dA = Qint / εₙ

let's form a Gaussian surface that is parallel to the surface, for example, a Cube. As the field is vertical and perpendicular to the surface, the field lines and the area vector are parallel whereby the scalar product is reduced to an ordinary product.

Φ = E A = Qint / ε₀

A = 1 km² (1000 m / 1km)² = 1 10⁶ m²

We can calculate the charge

Qint = E A ε₀

Qint = 81 1 10⁶ 8.85 10⁻¹²

Qint = 7.17 10⁻⁴ C

5 0

3 years ago

A pendulum has 968 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

Black_prince [1.1K]

968 j or kg because it cannot lose any of it energy and cannot gain any extra mass from nothing

8 0

3 years ago