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3 years ago

6

Formulating a Hypothesis: Part I Since the investigative question has two variables, you need to focus on each one separately. T

hinking only about the first part of the question, mass, what might be a hypothesis that would illustrate the relationship between mass and kinetic energy? Use the format of "if...then... because when writing your hypothesis
Help?

2 answers:

Bad White [126]3 years ago

4 0

Answer:

If the mass of an object increases, then its kinetic energy will increase proportionally because mass and kinetic energy have a linear relationship when graphed.

Explanation:

Westkost [7]3 years ago

3 0

Answer : The kinetic energy depends directly on the mass of a particle.

Explanation :

We know that the kinetic energy of any particle is given by :

$KE=\dfrac{1}{2}mv^2$

Where,

m is the mass of an object.

v is the velocity with which it is moving

Kinetic energy is due to the motion of the particle.

So, the kinetic energy of a particle is directly proportional to its mass.

Hence, the conclusion of the question is if the mass of a particle is increases then its kinetic energy also increase.

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Which situations might cause two observers (A and B) to measure different frequencies for the same vibrating object? Select the

Alex787 [66]

We want to explain why two different observes may measure different frequencies for the same vibrating object.

We will see that the two correct options are:

<em>Observer A is stationary and Observer B is moving.</em>
<em>Observer A and Observer B are moving at different speeds relative to each other.</em>

<em />

Let's assume that the vibrating object is a guitar string. Thus, the string makes a noise, and from that noise, we can estimate the frequency at which the string vibrates.

Now there appears a really cool effect, called the Doppler Effect. It says that the apparent change of frequency is <u>due to the motion of the observer or the source of the frequency (or both).</u>

For example, if you move towards the vibrating string, the perceived frequency will be larger, and you will hear a "higher" sound.

While if you move away from the string, the opposite happens, and you will hear a "lower" sound.

Then the only thing that impacts in how we perceive the frequency is our velocity relative to the source.

So, why do observers A and B measure different frequencies?

The two correct answers are:

<em>Observer A is stationary and Observer B is moving.</em>
<em>Observer A and Observer B are moving at different speeds relative to each other.</em>

If you want to learn more, you can read:

brainly.com/question/17107808

6 0

3 years ago

Which of the following are true for acceleration?

Fittoniya [83]

The SI unit for acceleration is m/s2 ( D)

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3 years ago

What is the oxidation state of a hydrogen atom bound to an iron atom.?

tresset_1 [31]

the answer is rust so the answer is rust

8 0

4 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

At the _______________ point, the particles in an object have not kinetic energy

Brums [2.3K]

The freezing point ..... :)

4 0

3 years ago