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1 year ago

Can somebody please help

Physics

1 answer:

8090 [49]1 year ago

5 0

The amplitude of the wave on the given sinusoidal wave graph is 10 cm.

<h3>What is amplitude of wave?</h3>

The amplitude of a wave is the maximum displacement of a wave. This is the highest vertical position of the wave from the origin.

Amplitude of the wave is calculated as follows;

From the graph, the amplitude of the wave or maximum displacement of the wave is 10 cm.

Thus, the amplitude of the wave on the given sinusoidal wave graph is 10 cm.

Learn more about wave amplitude here: brainly.com/question/25699025

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Two metra trains approach each other on separate but parallel tracks. one has a speed of 90 km/hr, the other 80 km/hr. initially

Gennadij [26K]

The trains take <u>57.4 s</u> to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

$v_A=90 km/h\\ v_B=-80 km/h$

The relative velocity of the train A with respect to B is given by,

$v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h$

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

$t= \frac{d}{v_A_B}$

Substitute 2.71 km for d and 170 km/h for $v_A_B$

$t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h$

Express the time in seconds.

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2 years ago

Unit 5 lesson 7 physical science 12 question

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Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

<u>Average Acceleration </u>

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that. </em>

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$

When $t_1=0\ sec$ , the object is at rest, $v_1=0$

We compute the average acceleration as

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

$\displaystyle a=\frac{36\ cm/s-0\ cm/s}{4\ sec-0\ sec}$

$\displaystyle a=\frac{36\ cm/s}{4\ s}$

$\boxed{a=9\ cm/s^2}$

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2 years ago

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