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2 years ago

Use words in the box to complete the diagram, and the definitions below it. Some words will be used

Physics

2 answers:

Masteriza [31]2 years ago

7 0

The correct words to fill the given blanks would be as follows:

- Peak

- Trough

- Equilibrium Line

- Oscillation

- Frequency, Hertz

- Period, seconds

The "peak" is denoted as the highest or maximum point that the wave attains while 'Trough' is characterized as the base of the wave.
The 'equilibrium line' is characterized as the extent to which the wave flattens or compresses.
Oscillation is defined as the process in which an object moves in the "back and forth motion."
Frequency is the amount of occurrences that waves travel in a second which is denoted in the terms of Hertz.
Lastly, period is the duration required by wave to accomplish one motion is denoted in terms of seconds.
Amplitude is the breadth or magnitude of the waves.

Learn more about "Oscillation" here:

brainly.com/question/9820984

Lostsunrise [7]2 years ago

3 0

Answer:

There ya go

Explanation:

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3 years ago

Read 2 more answers

What is true of an object pulled inward in an electric field?

slava [35]

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

So only option b is right

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3 years ago

Question 10 (1 point)

torisob [31]

Answer:

It does not hit the students face because the speed of the balloon slows down as energy is lost through thermal.

Explanation:

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3 years ago

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$