Two identical point charges of +Q coul are separated by a distance of 10 cm.

If an X-ray tube is operating at a current of 30.0 mA:a) How many electrons are striking the target per second? b) If the potent

Zina [86]

Answer:

Explanation:

A. Q= It

30E-3A=q

Ne = q

30E-3/1.6*10-19= N

N= 1.8*10^16 electrons

B. Power= I x v

= 30*10^-3A x 100*10^3v

= 3000watts

3 0

2 years ago

Consider an oblique shock wave with a wave angle of 35 o . Upstream of the wave, the static pressure and temperature are 2,000 l

ziro4ka [17]

Answer:

The pressure is 6570 lbf/ft²

The temperature is 766 ⁰R

The velocity is 2746.7 ft/s

deflection angle behind the wave is 17.56⁰

Explanation:

Speed of air at initial condition:

$a_1 = \sqrt{\gamma RT } = \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s$

γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.

initial mach number

$M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7} = 3$

then, $M_n = M_1sin \beta = 3sin(35) = 1.721$

based on the values obtained, read off the following from table;

P₂/P₁ = 3.285

T₂/T₁ = 1.473

Mₙ₂ = 0.6355

Thus;

P₂ = 3.285P₁ = 3.285(2000) = 6570 lbf/ft²

T₂ = 1.473T₁ = 1.473(520⁰R) = 766 ⁰R

Again; to determine the velocity and deflection angle, first we calculate the mach number.

$M_t_1 = M_1cos \beta = 3 cos(35) = 2.458$

$w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s$

$a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s$

$v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s$

$Tan(\beta -\theta) = \frac{v_2}{w_2} = \frac{862.59}{2746.7} \\\\Tan(\beta -\theta) = 0.314\\\\\beta -\theta= 17.44\\\\\theta = \beta - 17.44 = 17.56^o$

6 0

3 years ago

Convert 30.0 degrees Celsius into Fahrenheit

scoray [572]

The answer is 86 degrees Fahrenheit. Formula is (30 x 9.5) + 32 = 86

6 0

2 years ago

What type of circuit is illustrated?

Debora [2.8K]

Answer:

Parallel circuit

Explanation:

A parallel circuit is a closed circuit in which current flows and divide in two or more paths and recombining to complete the circuit, each load (light bulb) receives the fully voltage of the batteries in the circuit.

4 0

2 years ago

A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.

finlep [7]

Answer:

The acceleration expressed in the new units is $114.048 Km/h^{2}$

Explanation:

To convert from $m/s^{2}$ to $Km/h^{2}$ it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

$8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{3600s}{1h})^{2}$

$8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{12960000s^{2}}{1h^{2}})$

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

$114.048 Km/h^2$

So the acceleration expressed in the new units is $114.048 Km/h^2$ .

6 0

3 years ago