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aliina [53]
3 years ago
12

Julie is comparing electromagnetic waves and sound waves. Which of the following words would best complete her chart below? (2 p

oints)
Electromagnetic

Are transverse

Can travel _________

Fastest without a medium

Group of answer choices

without frequency

through any substance

on a curved path

in a vacuum
Physics
2 answers:
melisa1 [442]3 years ago
6 0

Answer:

D: In a vacuum

Explanation:

i took the test

elena55 [62]3 years ago
5 0

Answer:

in a vacuum

Explanation:

In physics, a transverse wave is a moving wave whose oscillations are perpendicular to the direction of the wave or path of propagation. In a transverse wave, the particles are displaced perpendicular to the direction the wave travels. Electromagnetic waves are transverse waves. They require no material medium for propagation and travel fastest in a vacuum.

Sound waves are longitudinal waves. In longitudinal waves, the vibrations are parallel to the direction of wave travel. Sound waves do not travel through vacuum because they require a material medium (air) for propagation. A vacuum has no air hence sound waves do not travel through a vacuum.

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The swinging pendulum has 10 joules of potential energy at its maximum height at points (1) and (5). If the mass of the pendulum
SVEN [57.7K]

The speed of the pendulum at point 3 is 1.4 m/s

Explanation:

We can solve this problem by using the law of conservation of energy. In fact, the mechanical energy of the pendulum (which is the sum of his potential energy + his kinetic energy) must be conserved. So we can write:

U_1 +K_1 = U_3 + K_3

where

U_1 is the initial potential energy, at the highest position

K_1 is the initial kinetic energy, at the highest position

U_3 is the final potential energy, at the lowest position

K_3 is the final kinetic energy, at the lowest position

We are told that:

U_1 = 10 J is the potential energy of the pendulum at the maximum height

K_1 = 0 (when the pendulum is at maximum height, the speed is zero, so the kinetic energy is zero)

U_3 = 0 (potential energy is zero at the lowest position)

Therefore,

K_3 = U_1 = 10 J

Kinetic energy can be rewritten as

K_3 = \frac{1}{2}mv^2

where

m = 10 kg is the mass of the pendulum

v is its speed at point 3

Solving for v,

v=\sqrt{\frac{2K_3}{m}}=\sqrt{\frac{2(10)}{10}}=1.4 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

4 0
3 years ago
What can happen to the motion of an object when there is a resultant force on it
MA_775_DIABLO [31]
Then, it would continue the motion in direction of resultant force...
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PLZ HELP DUE IN 10 MINUTES
likoan [24]

Answer:50%

Explanation: put it you are in a hurry

6 0
3 years ago
Why is alternating current more effective at long–distance travel than direct current?
amid [387]
Because we can send high voltage, low current which is more efficient since large amounts of current in the line causes power inefficiencies. AC can be transformed locally to high current lower voltage
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3 years ago
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A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta
Vika [28.1K]

Answer:

t_t=4.131\ s

Explanation:

Given:

height above the horizontal form where the ball is hit, y=1\ m

angle of projectile above the horizontal, \theta=30^{\circ}

initial speed of the projectile, u=40\ m.s^{-1}

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

u_y=u.\sin\theta

u_y=40\times \sin30^{\circ}

u_y=20\ m.s^{-1}

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

  • v_y=0\ m.s^{-1}

Using eq. of motion:

v_y^2=u_y^2-2g.h (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

0^2=20^2-2\times 9.8\times h

h=20.4082\ m (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

v_y=u_y-g.t

0=20-9.8t

t=2.041\ s

<u>Time taken to descent the total height:</u>

  • we've total height, h'=h+y =20.4082+1

h'=u_y'.t'+\frac{1}{2} g.t'^2

  • during the course of descend its initial vertical velocity is zero because it is at the top height, so u_y'=0\ m.s^{-1}

21.4082=0+4.9t'^2

t'=2.09\ s

<u>Now the total time taken by the ball to hit the ground:</u>

t_t=t'+t

t_t=2.09+2.041

t_t=4.131\ s

3 0
3 years ago
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