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beks73 [17]
2 years ago
13

When a satellite is in orbit around the Earth, the force of the gravity on the satellite is...

Physics
1 answer:
lbvjy [14]2 years ago
4 0

Answer:

Is always towards the center of the Earth

Explanation:

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Sound Waves can be used to measure ocean depth. This done by sending a sound wave to the floor of the ocean and recording the ti
vazorg [7]
When the sound wave returns to the machine, you can measure
how long it took to return. 

(You may notice that it's working just like RADAR, which does the
same thing with radio waves instead of sound waves.)

Even if you know how long the sound took to get to the bottom and
return to the top, you can't DO anything with this information if you
don't know the SPEED of the sound through the water.  Not only
the inventory of this machine, but anyone who uses it, has to know
the speed of the sound through water in order to use the round-trip
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3 0
3 years ago
(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the ca
gulaghasi [49]

Answer:

(a) E_{ c} = 112.5 \mu J

(b) E'_{ c} = 18 \mu J

Solution:

According to the question:

Capacitance, C = 1.00\mu F = 1.00\times 10^{- 6} F

Voltage of the battery, V_{b} = 15.0 V

(a)The Energy stored in the Capacitor is given by:

E_{c} = \frac{1}{2}CV_{b}^{2}

E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}

E_{ c} = 112.5 \mu J

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

E'_{c} = \frac{1}{2}CV'_{b}^{2}

E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}

E'_{ c} = 18 \mu J

6 0
3 years ago
According to the theory of plate tectonics
Darya [45]

Answer:

Alfred Wegener proposed the idea that earths lithospher is divided into tectonic plates.The theory was only accepted after he died.

8 0
3 years ago
Can you please solve this for me urgently want to make sure if my answers are correct?
MA_775_DIABLO [31]

Answer:

<u>Resolving</u><u> </u><u>horizontally</u><u>.</u> :

\sum d_{x} =  - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\  { \underline{d _{x} =  -  5.702 \: m}} \\  \\  \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0  \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y}  =  - 11.476 \: m}}

therefore, for resultant:

d =  \sqrt{ {d _{y} }^{2} + d _{x}  {}^{2}  }

substitute:

d =  \sqrt{ {( - 5.702)}^{2} +  {( - 11.476)}^{2}  }  \\  \\ d =  \sqrt{164.211}  \\  \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}

6 0
2 years ago
A 100 mH inductor whose windings have a resistance of 6.0 Ω is connected across a 12 Vbattery having an internal resistance of 3
Alexus [3.1K]

Answer:

The store energy in the inductor is 0.088 J

Explanation:

Given that,

Inductor = 100 mH

Resistance = 6.0 Ω

Voltage = 12 V

Internal resistance = 3.0 Ω

We need to calculate the current

Using ohm's law

V = IR

I=\dfrac{V}{R+r}

Put the value into the formula

I=\dfrac{12}{6.0+3.0}

I=1.33\ A

We need to calculate the store energy in the inductor

U=\dfrac{1}{2}LI^2

U=\dfrac{1}{2}\times100\times10^{-3}\times(1.33)^2

U=0.088\ J

Hence, The store energy in the inductor is 0.088 J

7 0
3 years ago
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