1) A star burst can be defined as stars formed from recycled dead star materials.

To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d

Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is $v = 350 \ m/s$

Explanation:

From the question we are told that

The distance of separation is d = 4.00 m

The distance of the listener to the center between the speakers is I = 5.00 m

The change in the distance of the speaker is by $k = 60 cm = 0.6 \ m$

The frequency of both speakers is $f = 700 \ Hz$

Generally the distance of the listener to the first speaker is mathematically represented as

$L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}$

$L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}$

$L_1 = 5.39 \ m$

Generally the distance of the listener to second speaker at its new position is

$L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}$

$L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}$

$L_2 = 5.64 \ m$

Generally the path difference between the speakers is mathematically represented as

$pD = L_2 - L_1 = \frac{n * \lambda}{2}$

Here $\lambda$ is the wavelength which is mathematically represented as

$\lambda = \frac{v}{f}$

=> $L_2 - L_1 = \frac{n * \frac{v}{f}}{2}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves

=> $5.64 - 5.39 = \frac{1 * v}{2*700}$

=> $v = 350 \ m/s$

7 0

2 years ago

A monoatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.010 m^3 to 0.040 m^3. The fina

Natali [406]

Answer:

A) 0.0 kJ

Explanation:

Change in the internal energy of the gas is a state function

which means it will not depends on the process but it will depends on the initial and final state

Also we know that internal energy is a function of temperature only

so here the process is given as isothermal process in which temperature will remain constant always

here we know that

$\Delta U = \frac{3}{2}nR\Delta T$

now for isothermal process since temperature change is zero

so change in internal energy must be ZERO

4 0

2 years ago

A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center

OlgaM077 [116]

Answer:

a) E = 8.63 10³ N /C, E = 7.49 10³ N/C

b) E= 0 N/C, E = 7.49 10³ N/C

Explanation:

a) For this exercise we can use Gauss's law

Ф = ∫ E. dA = $q_{int}$ /ε₀

We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product

The area of a sphere is

A = 4π r²

if we use the concept of density

ρ = q_{int} / V

q_{int} = ρ V

the volume of the sphere is

V = 4/3 π r³

we substitute

E 4π r² = ρ (4/3 π r³) /ε₀

E = ρ r / 3ε₀

the density is

ρ = Q / V

V = 4/3 π a³

E = Q 3 / (4π a³) r / 3ε₀

k = 1 / 4π ε₀

E = k Q r / a³

let's calculate

for r = 4.00cm = 0.04m

E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³

E = 8.63 10³ N / c

for r = 6.00 cm

in this case the gaussine surface is outside the sphere, so all the charge is inside

E (4π r²) = Q /ε₀

E = k q / r²

let's calculate

E = 8.99 10⁹ 3 10⁻⁹ / 0.06²

E = 7.49 10³ N/C

b) We repeat in calculation for a conducting sphere.

For r = 4 cm

In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.

E = 0

In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is

E = k q / r²

E = 7.49 10³ N / C

6 0

2 years ago

What is the tangential velocity at the edge of a disk of radius 10cm when it spins with a frequency of 10Hz? Give your answer wi

Nina [5.8K]

Answer:

630cm/s

Explanation:

In simple harmonic motion, the tangential velocity is expressed mathematically as v = ὦr

ὦ is the angular velocity = 2πf

r is the radius of the disk

f is the frequency

Given the radius of disk = 10cm

frequency = 10Hz

v = 2πfr

v = 2π×10×10

v = 200π

v = 628.32 cm/s

The tangential velocity = 630cm/s ( to 2 significant figures)

8 0

2 years ago

An object has a position given by r = [2.0 m + (2.00 m/s)t] i + [3.0 m − (1.00 m/s^2)t^2] j, where quantities are in SI units. W

lidiya [134]

Answer: 1 m/s

Explanation:

We have an object whose position $r$ is given by a vector, where the components X and Y are identified by the unit vectors $i$ and $j$ (where each unit vector is defined to have a magnitude of exactly one):