Which two characteristics of data always occur together?

A 50-V de voltage source was connected in series with a resistor and capacitor. Calculate the current in A to two significant fi

Naddik [55]

Answer:

Current through circuit will be $0.2706\times 10^6A$

Explanation:

We have given source voltage v = 50 volt

Resistance $R=25Mohm=25\times 10^6ohm$

Capacitance $C=0.1\mu F=0.1\times 10^{-6}F$

Time t = 5 sec

Time constant of RC circuit is given by $\tau =RC=25\times 10^6\times 0.1\times 10^{-6}=2.5sec$

We know that voltage across capacitor is given by $v_c=v_s(1-e^{\frac{-t}{\tau }})$

$v_c=50(1-e^{\frac{-5}{2.5 }})=43.2332v$

So current will be $=\frac{v_s-v_c}{R}=\frac{50-43.2332}{25\times 10^{-6}}=0.2706\times 10^6A$

So current through circuit will be $0.2706\times 10^6A$

4 0

4 years ago

What is kinematics? explain!! ~ thankyou ~

Novay_Z [31]

Answer:

Kinematics is the study of motion of a system of bodies without directly considering the forces or potential fields affecting the motion.

8 0

3 years ago

Read 2 more answers

We inhaie oxygen and exhale dioxide in the result of a what change

zzz [600]

Cellular respiration is the taking in of oxygen and release of the waste gas of carbon dioxide

8 0

3 years ago

At the same instant that a 0.50 kg ball is dropped from 25 m above Earth, a second ball, with a mass of .25 kg, is thrown straig

Anettt [7]

Answer:

A. 7.1m

B. 3.55m/s

C. 1.775m/s^2

Explanation:

First step is to identify given parameters;

Ball 1: m₁ = 0.5kg, u (initial velocity) =0, t = 2seconds

Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds

Second step: we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;

$y = ut - \frac{gt^2}{2}$

$y_{1} = 0 X 2 - \frac{9.8 X2^2}{2}$

$y_{1} = -19.6m$

Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)

[tex]y_{1} = 5.4m[/tex]

Third step: we determine the y-coordinate of ball 2 after 2 seconds

$y_{2} = 15 X 2 - \frac{9.8 X2^2}{2}$

$y_{2} = 10.4m$

Fourth step: we determine the y-component of the center mass of the two balls

$y = \frac{m_{1}y_{1} +m_{2}y_{2}}{m_{1} +m_{2} }$

$y = \frac{(0.5)X(5.4) +(0.25) X (10.4)}{(0.5 +0.25) }$

y = 7.1m

Fifth step: we solve B part of the question; velocity of the center mass of the two balls

$Velocity = \frac{distance of center mass of the two balls (y)}{time}$

$velocity = \frac{7.1 m}{2 s}$

velocity = 3.55m/s

Sixth step: we solve C part of the question; acceleration of the center mass of the two balls

$acceleration = \frac{velocity}{time}$

$acceleration = \frac{3.55}{2}$

acceleration = 1.775 m/s^2

7 0

3 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.6 cm and diameter 2.95 cm fr

slavikrds [6]

Answer:

48.4293354946 N

Yes

Explanation:

d = Diameter of rod = 2.95 cm

h = Length of rod = 92.6 cm

$\rho$ = Density of rod = 7800 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Volume of rod

$V=\dfrac{1}{4}\pi d^2h\\\Rightarrow V=\dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}$

Mass is given by

$m=\rho V\\\Rightarrow m=7800\times \dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}\\\Rightarrow m=4.93673144695\ kg$

Weight is given by

$W=mg\\\Rightarrow W=4.93673144695\times 9.81\\\Rightarrow W=48.4293354946\ N$

The weight of the rod is 48.4293354946 N

The mass of the rod is 4.93673144695 kg which is light. So, I will be able to carry the rod without a cart.

7 0

3 years ago