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BigorU [14]
2 years ago
9

What is the motion of the particles in this kind of wave?

Physics
2 answers:
MatroZZZ [7]2 years ago
8 0
I would say C but I’m not complete sore
xxTIMURxx [149]2 years ago
5 0

Answer:

A

Explanation:

A because its in large areas and there all spread apart more and there all going up and down.

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What is the relationship between the poles of a magnet? which direction does the magnet field move?
Assoli18 [71]
A magnet has a South Pole and a North Pole. South Pole and South Pole can't connect to her other, same as North and North. The same poles push each other away.
South Pole and North Pole connect.
4 0
3 years ago
a light-year is ____ A. the amount of time it takes light to get to the nearest star B. the distance light travels in a day C. t
Vinil7 [7]
The Answer is C. the distance light travels in a year
6 0
3 years ago
Read 2 more answers
An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result
mylen [45]

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

s=ut+\frac{1}{2}gt^{2}

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

   s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}

So the distance covered by the acorn is 3.136 m.

8 0
3 years ago
50200 J of heat are removed from
Dmitry_Shevchenko [17]

Correct Answer:

3.1375

Explanation:

Use equation Q=mcΔT to find m

Plug in all variables -50200=x\cdot 2000\cdot -8

Answer: 3.1375

4 0
3 years ago
g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor
Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1,  and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0
3 years ago
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