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Katarina [22]
3 years ago
6

Which two technologies use the same range of radio wave frequencies? television broadcasting and AM radio broadcasting cell phon

es and space communication CB radios and submarine communication AM and FM radio broadcasting
Physics
2 answers:
jolli1 [7]3 years ago
8 0

Answer:

The correct answer is cell phones and space communication.

Explanation:

I took the test on edge.

Tju [1.3M]3 years ago
4 0
Let us reject some answers. AM and FM radio waves do not interfere. Am are an orders of magnitude larger than FM waves. FM waves belong to microwaves, a subcategory of radiowaves while AM does not. Similarly, AM  and TV have not the same, because TV uses a subset of the frequencies used in FM. The correct answe is that CB radios and submarine communication happen at the same wavelengths/frequencies, namely low frequency radiowaves.
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Once an object enters orbit, what keeps the object moving sideways?
yaroslaw [1]

It's the natural tendency of things to keep going unless there's something trying to stop them.

It's usually called "inertia".

Don't get the idea from all of this that things stop unless there's something to keep them going.  The truth is exactly the opposite:  Things keep going unless there's something to make them stop.

7 0
3 years ago
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The lubrication of bone joints is a subject of ongoing medical research. Two bones connected at a joint do not touch. The bones
maks197457 [2]

The question is incomplete. The complete question is :

To measure the effective coefficient of friction in a bone joint, a healthy joint (and its immediate surroundings) can be removed from a fresh cadaver. The joint is inverted, and a weight is used to apply a downward force F⃗ d on the head of the femur into the hip socket. Then, a horizontal force F⃗ h is applied and increased in magnitude until the femur head rotates clockwise in the socket. The joint is mounted in such a way that F⃗ h will cause clockwise rotation, not straight-line motion to the right. The friction force will point in a direction to oppose this rotation.

Draw vectors indicating the normal force n⃗  (magnitude and direction) and the frictional force f⃗ f (direction only) acting on the femur head at point A.

Assume that the weight of the femur is negligible compared to the applied downward force.

Draw the vectors starting at the black dot. The location, orientation and relative length of the vectors will be graded

Solution :

The normal force represented by N is equal to the downward force, $F_d$ which is equal in magnitude but it is opposite in direction.

Also the frictional force acts always to oppose the motion because the bone starts moving in a clockwise direction. The frictional force that will be applied to the right direction so that the movement or the rotation at A is opposed.  

5 0
3 years ago
1. What is evaporation and how does it affect weather?
alina1380 [7]

Answer:

Explanation:don’t know

8 0
3 years ago
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball
dedylja [7]

Answer:

Magnitude of net force will be 432.758 N  

Explanation:

We have given x component of acceleration a_x=760m/sec^2

And vertical component of acceleration  a_y=770m/sec^2

Mass of the ball m = 0.40 kg

So net acceleration a=\sqrt{a_x^2+a_y^2}=\sqrt{760^2+770^2}=1081.896m/sec^2

Now according to second law of motion

Force = mass × acceleration

So F = 0.40×1081.896 = 432.758 N

3 0
3 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

5 0
3 years ago
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