What is the controlling factor for radio graphic contrast?

Which type of galaxy has the most active star formation?

UkoKoshka [18]

That answer is spiral galaxies

8 0

3 years ago

The inductive reactance of the circuit is exactly twice the resistance: XL=2R. Adjust the phasor that represents the voltage acr

SVETLANKA909090 [29]

Answer:

∅= $63.43^{0}$

Explanation:

z=impedance

$x_{l}$ =2R

R=R

The resultant of the resistances in the circuit is called impedance

$x_{l}$ is inductive reactance of the circuit

R is the resistance of the resistor

z= $\sqrt{xl^{2}+R^2 }$

z= $\sqrt{2R^{2}+R^2 }$

Z= $\sqrt{5R^2}$

Z=R $\sqrt{5}$ ohms

tan∅=2R/R

tan∅=2

∅=Tan^-1(2)

∅= $63.43^{0}$

phase angle is ∅= $63.43^{0}$

3 0

3 years ago

A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr

Ipatiy [6.2K]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = $\frac{9\times 10^9\times3\times10^{-9}}{d^2}$

Field due to charge = 4 X 10⁻⁹ C

$E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$

These two fields will be equal and opposite to make net field zero

$\frac{9\times 10^9\times3\times10^{-9}}{d^2}$ = $[tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$ [/tex]

$\frac{3}{d^2} =\frac{4}{(2-d)^2}$

$\frac{2-d}{d} =\frac{2}{1.732}$

d = 0.928

5 0

3 years ago

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x .........1

so force = mg = 0.35 (9.8) = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × $\sqrt{\frac{m}{k} }$ ................2

put here value

period of oscillations = 2π × $\sqrt{\frac{0.35}{28.58} }$

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0

3 years ago

A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the h

dybincka [34]

Given,

A player kicks a soccer hits at an angle of 30° at a speed of 26 m/s

We can resolute the trajectory of soccer into horizontal and vertical components.(Please see the attached file)

We can have,

Horizontal velocity component of ball= 26cos(30°) = 26×(√3÷2) = 22.51 m/s

And vertical velocity component of ball = 26sin(26°) = 26×(1÷2) = 13 m/s

6 0

3 years ago