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2 years ago

Subject- Science Grade- 6thplease help me with hw. Thank you!

Physics

2 answers:

Pavel [41]2 years ago

4 0

Answer:

Option b.

Hope it helps

velikii [3]2 years ago

3 0

Answer:

b. they get blown in from colder or warmer areas.

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An ac series circuit has an impedance of 60 Ohm and

Iteru [2.4K]

Answer:

Power factor of the AC series circuit is $cos\phi=0.5$

Explanation:

It is given that,

Impedance of the AC series circuit, Z = 60 ohms

Resistance of the AC series circuit, R = 30 ohms

We need to find the power factor of the circuit. It is given by :

$cos\phi=\dfrac{R}{Z}$

$cos\phi=\dfrac{30}{60}$

$cos\phi=\dfrac{1}{2}$

$cos\phi=0.5$

So, the power factor of the ac series circuit is $cos\phi=0.5$ . Hence, this is the required solution.

6 0

3 years ago

Any unit of acceleration MUST have the form of _____.

KengaRu [80]

Any unit of acceleration must have the dimensions (form) of

(a unit of length) / (a unit of time)²

7 0

3 years ago

HELP QUICK GIVING 16PTS!!!!!!! A student designs an experiment to test substances X, Y, and Z, to determine which one is a catal

Sunny_sXe [5.5K]

The test tube that first stops in bubbling or the production of a gas is the tube that contains the catalyst since the reaction ended faster than the others. A catalyst is known to speed up a reaction so it must the situation aforementioned is the answer.

5 0

3 years ago

Read 2 more answers

Please help i forgot how to solve this, thank you.

mixas84 [53]

Answer:

The radius of its orbit is $R \approx 1.899\times 10^{9}\,m$ .

Explanation:

Let suppose that Callisto rotates around Jupiter in a circular path and at constant speed, then we understand that net acceleration of this satellite is equal to the centripetal acceleration due to gravity of Jupiter. That is:

$\omega^{2}\cdot R = a_{net}$ (1)

Where:

$\omega$ - Angular speed, measured in radians per second.

$R$ - Radius of the orbit, measured in meters.

$a_{net}$ - Net acceleration, measured in meters per square second.

In addition, angular speed can be described in terms of period ( $T$ ), measured in seconds:

$\omega = \frac{2\pi}{T}$ (2)

And the net acceleration by the Newton's Law of Gravitation:

$a_{net} = \frac{G\cdot m}{R^{2}}$ (3)

Where:

$G$ - Gravitation constant, measured in cubic meters per kilogram-square second.

$m$ - Mass of Jupiter, measured in kilograms.

Now we apply (2) and (3) in (1) to derive an expression for the radius of the orbit:

$\frac{4\pi^{2}\cdot R}{T^{2}} = \frac{G\cdot m}{R^{2}}$

$R^{3} = \frac{G\cdot m \cdot T^{2}}{4\pi^{2}}$

$R = \sqrt[3]{\frac{G\cdot m\cdot T}{4\pi^{2}} }$ (4)

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $m = 1.90\times 10^{27}\,kg$ and $T = 1460160\,s$ , then the radius of the orbit of Callisto is: