If the moon were twice its present distance from earth, what would be the most noticeable effects on earth events

A liquid of density 1390 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the

Aloiza [94]

Answer:

The difference of power is

ΔP = 172.767 kPa

Explanation:

ρ = 1390 kg / m³

v = 9.63 m/s

d₁ = 10.1 cm , d₂ = 15.3 cm

Δz = 8.85 m

To find the difference ΔP between the fluid pressure at locations 2 and the fluid pressure at location 1

ΔP = ρ * g * Z + ¹/₂ * ρ * v² * ( 1 - (d₁ / d₂)⁴ )

ΔP = 1390 kg / m³ * 9.8 m/s² * 8.85 m + 0.5 * 1390 kg / m³ *(9.63 m /s)² * (1 - (0.101 m / 0.153 m )⁴ )

ΔP = 172.767 x 10 ³ Pa

ΔP = 172.767 kPa

6 0

3 years ago

- A capacitor's dielectric material is a vacuum. The capacitor's dielectric constant, or K, will be equal to

dlinn [17]

If a capacitor's dielectric constant is vacuum its dielectric constant is k will be equal to 1.

<u>Explanation:</u>

Relative permittivity of a dielectric substance is referred to as its dielectric constant. Relative permittivity/dielectric constant k is a dimensionless quantity that is the ratio of absolute permittivity and vacuum permittivity.

It is given by the expression

k=k=ε /ε0

where ε denotes absolute permittivity and ε0 denotes permittivity of vacuuum.

Absolute permittivity ε of vacuum= ε0

therefore k= ε0/ ε0=1

dielectric constant of vacuum is 1 .

4 0

3 years ago

A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle du

White raven [17]

Answer:

$a. \phi _B=3.0528\times10^-^3T\ m^2\\ \\b. \phi_B=1.83299\times10^-^3T\ m^2\\\\c.\phi_B=0$

Explanation:

#Consider a circular area of radius $R=2.98cm$ in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

a. first, find the magnetic flux if the magnetic field has a magnitude of $B=0.23T$ and points toward the +ve z-axis. The angle between the magnetic field and the area is $\theta=0$ . Hence the magnetic flux:-

$\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2$

Hence flux magnitude in $+z$ direction is $3.0528\times10^-^3T \ m^2$

b. We now find the magnetic flux when the field has a magnitude of <em>B=0.230T</em> and points at an angle of $\theta=53.1\textdegree$ from the $+z$ direction.

Magnetic flux is calculated as:

$\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2$

Hence the flux at an angle of $53.1\textdegree$ is $1.83299\times 10^-^3T \ m^2$

c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

$\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0$