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Akimi4 [234]
3 years ago
7

When demand for electric power is high, power company sometimes reduced the voltage, thereby producing a "brown-out." What is be

ing saved?
Physics
1 answer:
Bess [88]3 years ago
8 0
They sometimes reduce the voltage to prevent a potential blackout caused by the system being overworked, this saves them money that they would have to spend to make repairs
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Which statements are true about galaxies, stars, and the universe? (more than one true answer by the way)
Elina [12.6K]

Answer:

statements <em><u>2, 3, 4, and 7</u></em> are true

Explanation:

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3 years ago
Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi
Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite r_{1}= 8.20\times10^{7}

Orbital radius of second satellite r_{2}=7.00\times10^{7}\ m

Mass of first satellite m_{1}=53.0\ kg

Mass of second satellite m_{2}=54.0\ kg

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

v=\sqrt{\dfrac{GM}{r}}

From this relation,

v_{1}\propto\dfrac{1}{\sqrt{r}}

Now, \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}

v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}

Put the value into the formula

v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}

v_{2}=5195.16\ m/s

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0
3 years ago
Finderkeepers,ismadeormodel
ankoles [38]

Answer:

model lol

Explanation:

7 0
3 years ago
Speakers A and B are vibrating in phase. They are directly facing each other, are 8.2 m apart, and are each playing a 78.0 Hz to
Stels [109]

Answer:6.298,4.1,1.9015

Explanation:

Wavelength=\frac{velocity of sound }{frequency}

=\frac{343}{78}=4.397 m

Distance of 3rd speaker from speaker A is x

From B 78-x

Difference between the distances must be a whole number of wavelengths

First

x-\left ( 8.2-x\right )=4.397    for 1 st wavelength

2x=8.2+4.397=12.597

x=6.298m

second

For zero wavelength

x-\left ( 8.2-x\right )=0

2x=8.2

x=4.1m

Third

\left ( 8.2-x\right )-x=4.397

x=1.9015 m

6 0
3 years ago
A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (
SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

#SPJ1

4 0
2 years ago
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