Through which material do P waves move the fastest?

The following are ways to properly manage your stress except

ElenaW [278]

Answer:

you should provide the options for us to answer?

7 0

3 years ago

George rides his bike to his friend’s house that is 5 kilometers from his house. If he rides his bike at an average speed of 15

oksano4ka [1.4K]

I haven’t learned math in forever sorry

7 0

3 years ago

consider two stars, star a and star b. star a has a temperature of 4900 k , and star b has a temperature of 9900 k . how many ti

ValentinkaMS [17]

The Energy flux from Star B is 16 times of the energy flux from Star A.

We have Two stars - A and B with 4900 k and 9900 k surface temperatures.

We have to determine how many times larger is the energy flux from Star B compared to the energy flux from Star A.

<h3>State Stephen's Law?</h3>

Stephens law states that if E is the energy radiated away from the star in the form of electromagnetic radiation, T is the surface temperature of the star, and σ is a constant known as the Stephan-Boltzmann constant then-

$$\frac{Energy}{Area} = \sigma\times T^{4}$

Now -

Energy emitted per unit surface area of Star is called Energy flux. Let us denote it by E. Then -

$$E= \sigma\times T^{4}$

Now -

For Star A →

$T_{A}$ = 4900 K

For Star B →

$T_{B}$ = 9900 K

Therefore -

$$\frac{T_{B} }{T_{A} } =\frac{9900}{4900}$

$\frac{T_{B} }{T_{A} }=$ 2.02 = 2 (Approx.)

Now -

Assume that the energy flux of Star A is E(A) and that of Star B is E(B). Then -

$$\frac{E(B)}{E(A)} = \frac{\sigma\times T(B)^{4} }{\sigma \times T(A)^{2} }$

E(B) = E(A) x $(\frac{T(B)}{T(A)} )^{4}$

E(B) = E(A) x $2^{4}$

E(B) = 16 E(A)

Hence, the Energy flux from Star B is 16 times of the energy flux from Star A.

To learn more about Stars, visit the link below-

brainly.com/question/13451162

#SPJ4

4 0

2 years ago

A 0.45 m radius, 500 turn coil is rotated one-fourth of a revolution in 4.01 ms, originally having its plane perpendicular to a

AysviL [449]

Answer:

B = 0.126 T

Explanation:

As per Faraday's law we know that rate of change in magnetic flux will induce EMF in the coil

So here we can say that EMF induced in the coil is given as

$EMF = \frac{\phi_2 - \phi_1}{\Delta t}$

initially the coil area is perpendicular to the magnetic field

and after one fourth rotation of coil the area vector of coil will be turned by 90 degree

so we can say

$\phi_1 = NBAcos 0 = BA$

$\phi_2 = NBAcos 90 = 0$

now we will have

$EMF = \frac{NBA}{t}$

$10,000 V = \frac{(500)(B)(\pi \times 0.45^2)}{4.01\times 10^{-3}}$

$B = 0.126 T$

3 0

3 years ago

g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller

levacccp [35]

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

$A_1v_1 = A_2v_2$

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

$(0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}$

v₂ = 30 m/s

b)

$m_1 = \rho A_1v_1t$

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

$m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\$

m₁ = 12600 kg

c)

$m_1 = \rho A_1v_1t$

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

$m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\$

m₂ = 12600 kg

7 0

3 years ago