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3 years ago

6

PLEASE HELP!!!!!! the independent variable is associated mainly with? A. control and experimental groupsB. the treatment that th

e researcher controls directly C. the result of experiment D. the hypothesis

2 answers:

Alex_Xolod [135]3 years ago

7 0

The correct answer is B. The independent variable refers to what is being tested by a researcher in an experiment

Hope this helps

aliina [53]3 years ago

6 0

Answer:

B

Explanation:

Apex :)

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A 0.74 mF capacitor is connected to a standard outlet (rms voltage 82 V, frequency 49 Hz ). Determine the magnitude of the curre

disa [49]

Answer:

I = 26.36 cosω t A

Explanation:

Given that

C=0.74 mF

Vrms= 82 V

Frequency ,f= 49 Hz

We know that ω = 2 π f

ω = 2 x π x 49

ω = 307.72 rad/s

As we know that voltage given as

V= Vo sinω t

$V_o=\sqrt2\ V_{rms}$

$V_o=\sqrt2\ \times 82$ \

Vo=115.96 V

V=115.96 sinω t

The current given as

$I=C\dfrac{dV}{dt}$

$I=0.74\times \dfrac{dV}{dt}\ mA$

$\dfrac{dV}{dt}=115.96\omega cos\omega t$

$I=0.74\times 115.96\times 307.22 cos\omega t\ mA$

I = 26362.67 cosω t mA

I = 26.36 cosω t A

This is the current at time ant time t.

7 0

3 years ago

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

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3 years ago

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

2 years ago

Two identical objects are moving directly toward one another at the same speed v. ~v −~v m m What is the total kinetic energy of

Natalka [10]

<span>Answer: Correct answer is just add the two kinetic energies; E = (1/2)mv^2 + (1/2)mv^2</span>

5 0

2 years ago

Read 2 more answers

If someone is driving 100 miles in 60 minutes then drives 150 miles in 100 minutes west, what is his acceleration rate.

Sliva [168]

Answer:

his acceleration rate is -0.00186 m/s²

Explanation:

Given;

initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)

time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s

final position of the car, x₁ = 150 miles = 241,350 m

time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s

The initial velocity is calculated as;

u = 160, 900 m / 3,600 s

u = 44.694 m/s

The final velocity is calculated as;

v = 241,350 m / 6,000 s

v = 40.225 m/s

The acceleration is calculated as;

$a = \frac{\Delta V}{\Delta t} = \frac{v- u}{t_1 - t_ 0} = \frac{40.225 - 44.694}{6000-3600} = -0.00186 \ m/s^2\\\\$

Therefore, his acceleration rate is -0.00186 m/s²

6 0

3 years ago