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3 years ago

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What would a scientist most likely do to demonstrate the effects of a tsunami? (2 points) construct tsunami models make underwat

er plates move create actual underwater earthquakes take underwater photographs during a tsunami

1 answer:

Otrada [13]3 years ago

5 0

Construct tsunami models

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You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-hi

Andrew [12]

Answer:

22.15 N/m

Explanation:

As we know potential energy = m*g*h

Potential energy of spring = (1/2)kx^2

m*g*h = (1/2)kx^2

Substituting the given values, we get -

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

k = 39200/2.645

k = 19600 N/m

For safety reasons, this spring constant is increased by 13 % So the new spring constant is

k = 19600 * 1.13 = 22148 N/m = 22.15 N/m

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3 years ago

Two identical balloons of the same volume are pumped up with air to more than atmospheric pressure and suspended on the ends of

Umnica [9.8K]

Answer:

The balance will be upset and it will tip to the side of the still pumped ballon.

Explanation:

Given that the balloons are pumped to more than atmospheric pressure it means that the air inside has a higher density than the air outside. If you punture one balloon the air will be released and now you have on both sides the weight of the balloons itselves, but on the side of the still pumped balloon you will have a greater amount of air (is more dense) so more weight. Considering this, the balance will be upset and it will tip to the side of the still pumped ballon.

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4 years ago

Destiny can u come comment on this plz

goblinko [34]

Answer:

Where's Destiny?

When's Destiny?

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4 years ago

A string is strung horizontally with a fixed tension. A wave of frequency 90 Hz is sent along the string, and it has a wave spee

ankoles [38]

Answer:

wave speed of the second wave 55.1 m/s

Explanation:

The wave speed of the second wave will going to be same i.e 55.1 m/s. As speed of wave is dependent on the medium and here in this case medium is remain same for second wave also i.e string only, no change in the medium noticed, therefore speed of second wave still remain same i.e 55.1 m/s.

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3 years ago

A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin

pychu [463]

a) See graph in attachment

b) The suvat equation to use is $v_f^2 - v_i^2 = 2as$

c) The acceleration is $a=\frac{v_f^2-v_i^2}{2s}$

d) The acceleration is $1.25 m/s^2$

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is $x_i = 0$ , the origin

- The final position of the boat is $x_f = 200 m$

- The initial velocity of the boat is $v_i = 20.0 m/s$

- The final velocity of the boat is $v_f = 30.0 m/s$

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

$v_f^2 - v_i^2 = 2as$

where

$a$ is the acceleration

c)

Now we have to solve the equation

$v_f^2 - v_i^2 = 2as$

In order to find the acceleration.

This can be done by dividing both terms by $2s$ : this way, we find

$\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}$

And so the acceleration is

$a=\frac{v_f^2-v_i^2}{2s}$

d)

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

And substituting into the equation,

$a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2$

e)

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

$v_f = v_i + at$

where:

$v_i$ is the initial velocity

$v_f$ is the final velocity

a is the acceleration

t is the time

Here we have:

$v_i = 20.0 m/s$

$v_f = 30.0 m/s$

$a=1.25 m/s^2$

Solving for t, we find:

$t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s$

Learn more about accelerated motion:

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4 years ago