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8090 [49]
2 years ago
12

Texting: what are two needs or wants that this wave technology meets it:

Physics
1 answer:
qaws [65]2 years ago
3 0

Answer:

bugs/viruses all the time and more social platforms

Explanation:

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the standard way the body is positioned when using anatomical terminology ... invisible line that runs vertically through the center of the axial region.

Explanation:

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2 years ago
What is the gravitational potential energy of a tiger that weighs 200 N standing on a rock that is 2 m above the ground?
Nikitich [7]
G.P.E = mgh
Weight = mg = 200N

So G.P.E = 200 * 2 = 400 Joules
5 0
3 years ago
Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second
Anit [1.1K]
The amplitude of a wave can be obtained by measuring the distance from the resting position of the wave to its crest. The resting position is half of the distance from the crest to the trough. Given that the distance between the crest and the trough is 3 meters, the amplitude should be half of that, which is 1.5 meters.
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3 years ago
A train travels 90 kilometers in 4 hours, and then 82 kilometers in 1 hours. What is its average speed?
kvasek [131]
Average speed = total distance / total time.

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5 0
2 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as m_{1} \ and\ m_{2}\ respectively

Let\ m_{1} =1 kg\ and\ m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [\mu]=0.13

Hence\ the\ frictional\ force\ F=\mu N

                                                     F=\mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-\mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-\mu m_{2}g= m_{2} a   [2]

                           ---------------------------------------------------

                           [m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g

                           a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2

                           a=\frac{0.74}{3} *9.8\ m/s^2

                           a=2.417 m/s         [ans]

6 0
3 years ago
Read 2 more answers
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