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3 years ago

50 points!

1 answer:

5 0

From Kepler's 3rd law, time period of satellite orbiting around planet in a circular orbit is given by,
T² = 4π²r³/GM
M= mass of the Earth = 6×10²⁴ Kg.
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴
T² = 1653339719
∴ T = 40661.28 seconds.

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An ideal monatomic gas expands isobarically, doubling the volume, from a state A to state B. It is then compressed isothermally

drek231 [11]

B because I just think it’s the right answer because drawing a PV diagram for this cycle is the best answer even though it might be a stupid so I would not pay attention to my response because I’m just getting this for points but let me know if it is the right answer because right now I’m just guessing.

8 0

3 years ago

HELP PLEASE!!!

Andru [333]

Answer:

It is si-32

Explanation:

5 0

3 years ago

Please help with Physics Circuits!

Zigmanuir [339]

1) Let's start by calculating the equivalent resistance of the three resistors in parallel, $R_2, R_3, R_4$ :
$\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}$
From which we find
$R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
$R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega$
So, the correct answer is D)

2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$

And these are connected in series with a resistor of $10 \Omega$ , so the equivalent resistance of the circuit is
$R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega$

And by using Ohm's law we find the current in the circuit:
$I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A$
So, the correct answer is C).

3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}$
From which we find
$R_{23} = 2.5 \Omega$
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
$R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega$

And by using Ohm's law we find the current flowing in the circuit:
$I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A$

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
$V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V$
So, the correct answer is D).

4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
$\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}$
From which we find
$R_{23} = 7.55 \Omega$

Now all the resistors are in series, so the equivalent resistance of the circuit is:
$R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega$

The current in the circuit is given by Ohm's law
$I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A$

Now we can compare the voltage drops across the resistors. Resistor 1:
$V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V$
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
$V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V$
Resistor 4:
$V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V$

So, the greatest voltage drop is on resistor 1, so the correct answer is D).

5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
$P_3 = I_3^2 R_3 = \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W$
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.

5 0

3 years ago

Read 2 more answers

Darina [25.2K]

Answer:

7.5 x 10⁻⁸N

Explanation:

Given parameters:

Mass 1 = 60kg

Mass 2 = 75kg

Distance between the bodies = 2m

Unknown:

Gravitational fore = ?

Solution:

The gravitational force between the two bodies can be derived using;

F = $\frac{G mass 1 x mass 2}{distance^{2} }$

G is the universal gravitation constant = 6.67 x 10⁻¹¹m³kg⁻¹s⁻²

Insert the parameters and solve;

F = $\frac{6.67 x 10^{-11} x 60 x 75}{2^{2} }$ = 7.5 x 10⁻⁸N

3 0

3 years ago

A group of skydiver are riding in a helicopter up to the spot form which they will jump. As they ride upwards. their gravitation

Dafna1 [17]

A group of skydiver are riding in a helicopter up to the spot form which they will jump. As they ride upwards. their gravitational potential energy increases.

Answer: Option A

<u>Explanation:</u>

Gravitational potential energy refers exerted energy on any object rising from the surface of Earth. It is the measurement of potential energy of any object depending on its position.

We can also define gravitational potential energy as the amount of energy required to raise an object to certain level from the surface of Earth or the work done to raise the object to some position from a position of reference. The formula for calculating the gravitational potential energy is

$\text {Gravitational potential energy}=m g \Delta h$

Here, m is the object's mass, g -the acceleration and $\Delta h$ - the object's distance from the Earth's surface.

So from this formula, it can be confirmed that if the helicopter carrying skydivers are made to ride upwards, then their gravitational potential energy will increase as the distance of helicopter from the Earth's surface is increased on riding the helicopter upwards.

6 0

3 years ago