A ball is dropped from a height of 2 m. how long will it take this ball to reach the ground?

Manufacturing Plant Power A manufacturing plant uses 2.36 kW of electric power provided by a 50.0 Hz ac generator with an rms vo

Pie

Answer:

(a) Z = 48.3 Ω

(b) cos ∅ = 0.455

(c) Irms = 10.35 A

(d) C = 74.02 μF

(e) Irms = 4.44 A

Explanation:

Power (P) = 2.36 kW

Frequency (f) = 50 Hz

RMS Voltage (Vrms) = 500 V

Resistance (R) = 22 Ω

Inductive Reactance (XL) = 43 Ω

(a) to calculate the total impedance, use the formula:

Z = √(R² + XL²)

= √((22)² + (43)²)

= √2333

Z = 48.3 Ω

(b) To calculate the plant's power factor, we will use the formula:

cos ∅ = R/Z

= 22/48.3

cos ∅ = 0.455

(c) To calculate the RMS current used by the plant, divide the RMS voltage value by the impedance of the plant.

Irms = Vrms/Z

= 500/48.3

Irms = 10.35 A

(d) For the power factor to become unity, the inductive reactance must be equal to the capacitive reactance i.e. Xc = XL

Xc = XL

1/(2πfC) = XL

1/(2πfXL) = C

C = 1/(2π*50*43)

= 7.402 x 10⁻⁵

C = 74.02 μF

(e) P = Vrms*Irms*cos∅

Irms = P/Vrms*cos∅

= 2.22 x 10³/500*1

Irms = 4.44 A

4 0

3 years ago

A stone tumbles into a mine shaft and strikes bottom after falling for 4.2 second how deep is the mine shaft

rjkz [21]

$4.2*9.8\\41.16$

41.16 meters

5 0

3 years ago

Explain how you can cause light separated by a prism to combine

r-ruslan [8.4K]

When you shine a lite through a prism is reflects out light through all of the edges and causes light separation. Or just simply shine a laser through the edge of a sideways piece of glass.

I hope that this was helpful for you.

7 0

3 years ago

A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1

nasty-shy [4]

(a) $1.72\cdot 10^{-5} \Omega m$

The resistance of the rod is given by:

$R=\rho \frac{L}{A}$ (1)

where

$\rho$ is the material resistivity

L = 1.20 m is the length of the rod

A is the cross-sectional area

The radius of the rod is half the diameter: $r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m$ , so the cross-sectional area is

$A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2$

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

- The voltage at this temperature is V = 15.0 V

- The current at this temperature is I = 18.6 A

So, the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega$

And now we can re-arrange the eq.(1) to solve for the resistivity:

$\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m$

(b) $8.57\cdot 10^{-4} /{\circ}C$

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

$R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega$

The equation that gives the change in resistance as a function of the temperature is

$R(T)=R_0 (1+\alpha(T-T_0))$

where

$R(T)=0.86 \Omega$ is the resistance at the new temperature (92.0°C)

$R_0=0.81 \Omega$ is the resistance at the original temperature (20.0°C)

$\alpha$ is the temperature coefficient of resistivity

$T=92^{\circ}C$

$T_0 = 20^{\circ}$

Solving the formula for $\alpha$ , we find

$\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C$

5 0

2 years ago

4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3

Kazeer [188]

Expand each vector into their component forms:

$\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N$

Similarly,

$\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N$

$\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N$

Then assuming the resultant vector $\vec R$ is the sum of these three vectors, we have

$\vec R=\vec A+\vec B+\vec C$

$\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N$

and so $\vec R$ has magnitude

$\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N$

and direction $\theta_R$ such that

$\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ$

5 0

3 years ago