All categories

3 years ago

PLEASE HELP ITS RLLY URGENT PLEASE:((((

Physics

2 answers:

SSSSS [86.1K]3 years ago

8 0

Answer:

8. Man-made object that orbits Earth

9. Red Giant

10. Super Red Giant

Sergio [31]3 years ago

4 0

Answer:

1. A satellite is an object which has been sent into space in order to collect information or to be part of a communications system. Satellites move continually round the Earth or around another planet.

2. red giant

3.red giant.

Explanation:

You might be interested in

Is it possible for a gas contained in a chamber to maintain a constant temperature while heat is being added to the gas? Explain

Evgesh-ka [11]

Answer:Yes

Explanation:

Yes it is possible for a gas contained in a chamber to maintain a constant temperature while heat is being added to the gas.A process in which temperature of the gas remains constant is called Isothermal Process.For an ideal internal energy is a function of temperature therefore internal energy remains constant while all the heat added is converted to do the work done by the system.

6 0

3 years ago

In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.

garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

$\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...$

Find the equivalent resistance of the right branch of the circuit:

$R_{eq} = R_{2} +R_{3} \\R_{eq} = 20.7 + 25.8 = 46.5 ohms$

$\frac{1}{R_{eq} } = \frac{1}{19.2} +\frac{1}{46.5}\\\\\frac{1}{R_{eq} } = 0.0735887097\\\\R_{eq} = 13.5890411$

6 0

2 years ago

An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made fro

Dmitrij [34]

Answer:

(a) 17634.24 Ω

(b) 0.0068 A

Explanation:

(a)

The formula for inductive inductance is given as

X' = 2πFL................... Equation 1

Where X' = inductive reactance, F = frequency, L = inductance

Given: F = 60 Hz, L = 46.8 H, π = 3.14

Substitute into equation 1

X' = 2(3.14)(60)(46.8)

X' = 17634.24 Ω

(b)

From Ohm's law,

Vrms = X'Irms

Where Vrms = Rms Voltage, Irms = rms Current.

make Irms the subject of the equation

Irms = Vrms/X'...................... Equation 2

Given: Vrms = 120 V, X' = 17634.24 Ω

Substitute into equation 2

Irms = 120/17634.24

Irms = 0.0068 A

5 0

3 years ago

An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete

Ksju [112]

<span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:

R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s

the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu! :)</span>

6 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

2 years ago