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Dmitry_Shevchenko [17]
3 years ago
10

Stars and planets form when dust and gas from space is pulled together by.. what?

Physics
1 answer:
NemiM [27]3 years ago
5 0

Answer:

gravity?

Explanation:

i dont have a lot of context but this is what i thought of so...

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g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie
polet [3.4K]

Answer: 71.16\ Hz

Explanation:

Given

Capacitance C=100\ \mu F

Resistance R=500\ \Omega

Inductance L=50\ mH

In LCR circuit, current is maximum at resonance frequency i.e.

X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}

Insert the values

\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}

Also, frequency is given by

\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}

\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz

8 0
3 years ago
24) Which of the following equations is correctly balanced? HELP ME ASAP PLEASE
hammer [34]

Answer:

the correct answer is option C

7 0
2 years ago
Read 2 more answers
An office building has a 24-volt branch circuit installed for landscape lighting around the front of the building. The circuit w
Arturiano [62]

The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

<h3>UF cable</h3>

UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.

For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.

Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

Learn more about UF cable here: brainly.com/question/8591560

5 0
2 years ago
a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity
stich3 [128]

consider the motion along the X-direction

X = horizontal displacement = 80 m

V_{ox} = initial velocity along the x-direction = v Cos60

t = time of travel

using the equation

X = V_{ox}   t

80 = (v Cos60) (t)

t = 160/v                                         eq-1


consider the motion in vertical direction :

Y = vertical displacement = 20 m

V_{oy}  = initial velocity in Y-direction = v Sin60

a = acceleration = - 9.8 m/s²

t = time of travel = 160/v

using the equation

Y = V_{oy}  t + (0.5) a t²

20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²

v = 32.5 m/s

4 0
3 years ago
-Plz Help Meh-<br> Calculate the acceleration of a car if the force is 450N and the mass is 1300kg.
Allisa [31]

Answer:

Explanation:

F=Ma

450=1300×a

Ans=0.346m/s²

4 0
3 years ago
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