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4 years ago

Whaaaaaaat????? I'm in the pit of dispare! Please help

Physics

2 answers:

sdas [7]4 years ago

3 0

The correct answer is c. Cirrus clouds.

alukav5142 [94]4 years ago

3 0

Hello there Matya

the answer is C.Cirrus clouds

Cirrus clouds are usually white and predict fair to pleasant weather.

hope this helps

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PLEASE HELP!!! FIRST IS BRAINLIEST!!!

Verizon [17]

1. A Medium

2. One of these 4, the refractive index of the medium <span>Choose whichever are on the test.

3. </span>properties and nature of medium, Hope this helps.

6 0

4 years ago

Read 2 more answers

Three electromagnetic waves travel through a certain point P along an x axis. They are polarized parallel to a y axis, with the

chubhunter [2.5K]

Answer:

$4.846*10^{-5} sin(5*10^{14}t )$

Explanation:

Frequency of each electromagnetic wave is same thus we can interfere them along X axis in Phase diagram.

In phase diagram indicate each wave as vector with amplitude as length and phase difference as angle. (shown in attachment)

Now take component of E2 and E3 amplitude along X direction.

Net amplitude of all E in X direction = $4*10^{-5}$ + cos(45)[ $6*10^{-6}$ + $6*10^{-6}$ ]

⇒ $4.846*10^{-5}$

Final resultant have no change in frequency

So , resultant is $4.846*10^{-5} sin(5*10^{14}t )$

6 0

4 years ago

The position coordinate of a particle which is confined to move along a straight line is given by s =2t3−24t+6, where s is measu

Gre4nikov [31]

Answer:

a) the answer is t=4 seconds

b) acceleration is zero

c) displacement= 142 m

Explanation:

Given the position of the particle

$s=2t^3-24t+6$

a) the time required when velocity $v=72 m/s$

$v=72=\frac{ds}{st}=6t^2-24$

now we solve for time $t$

$6t^2=72+24=96\\\\\implies t^2=\frac{96}{6}\\\therefore t=4 s$

b) acceleration when $v=30 m/s$

acceleration is the time derivative of velocity i.e

$a=\frac{dv}{dt}=\frac{d}{dt}(30)=0$

c) the net displacement of the particle during the interval t = 1 s to t = 4 s is

$s_4-s_1=2t^3-24+6|_{t=4}-2t^3-24+6|_{t=1}\\s_4-s_1=126-(-16)=142 m$

7 0

4 years ago

The first hill of a roller coaster is 42 meters high. The top of the second hill of the one

Marta_Voda [28]

Answer:

Part a)

Speed of the roller coaster is

$v = 13.3 m/s$

Part b)

Since it is moving with non zero speed at some height above the ground

So we will have

Kinetic energy + Potential energy Both

Explanation:

As we know that there is no friction on the path

So here we can use mechanical energy conservation law

so we will have

Part a)

$U_i + K_i = U_f + K_f$

$mgh + 0 = \frac{1}{2}mv^2 + mgH$

$34500(9.81)(42) = \frac{1}{2}(34500)v^2 + 34500(9.81)(33)$

$v = 13.3 m/s$

Part b)

Since it is moving with non zero speed at some height above the ground

So we will have

Kinetic energy + Potential energy Both

7 0

4 years ago

Read 2 more answers

Plz help fast!!

barxatty [35]

We are asked to compare two series circuits having equal number of light bulbs.
1st circuit is powered by 6 batteries each having a voltage of 1.5V
2nd circuit is powered by a single battery having a voltage of 9V.
The six batteries in the 1st circuit can be connected together in series or in parallel.
When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means
Voltage of pack = number of batteries*voltage of each battery
Voltage of pack = 6*1.5
Voltage of pack = 9 volts
But the current remains same in the series connection since there is only path for the current to flow.
On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.
Circuit 1:
In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Circuit 2:
In this circuit, we have 1 battery which provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit just like in circuit 1.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Conclusion:
Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

5 0

4 years ago