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3 years ago

Unlike an observational study, an experiment can prove:

Physics

1 answer:

Zinaida [17]3 years ago

3 0

A. Cause and effect.

Explanation:

An experiment can prove cause and effect in a study unlike observational studies.

Observational studies is a non-experimental system of investigation.
It helps to identify the causes(independent variable) and the effect(dependent variables).
In observational studies, the variables cannot be controlled and so it is difficult to prove the cause and effect.
An experiment can prove cause and effect because the parameters are tested and can be controlled.

Learn more:

Experiment brainly.com/question/5096428

#learnwithBrainly

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If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose

Volgvan

Answer:

a) A=0.125 m

b) T = 1.72 s

c) f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz

8 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.

Neporo4naja [7]

Answer:

The distance is $r_2 = 0.24 \ m$

Explanation:

From the question we are told that

The distance from the conversation is $r_1 = 24.0 \ m$

The intensity of the sound at your position is $\beta _1 = 40 dB$

The intensity at the sound at the new position is $\beta_2 = 80.0dB$

Generally the intensity in decibel is is mathematically represented as

$\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is also mathematically represented as

$d = \frac{P}{A}$

$\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=> $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=> $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=> $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

$P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

Now the power of the sound wave at the second position is mathematically represented as

$P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally power of the wave is constant at both positions so

$A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

$4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

$r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

substituting value

$r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

$r_2 = 0.24 \ m$

7 0

3 years ago

What is the energy of a photon with a frequency of 3. 6 Ă— 1015 Hz? Planckâ€™s constant is 6. 63 Ă— 10â€""34 Jâ€˘s. 1. 8 Ă— 10â€

erastovalidia [21]

Every photon has some energy within it and that energy is called photon energy.

The energy of photon is $2.38\times10^{-18}\;\rm J$ .

Given that, frequency of the photon is $3.6\times10^{15}\;\rm Hz$ and Planck's constant is $6.626\times {{10}^{-34}}$ .

So the energy of the photon can be calculated by the given formula.

$E=h\nu$

Where $h$ is plank's constant and $\nu$ is the frequency of the photon.

By substituting the values in the above formula, the photon energy is,

$E=6.626\times10^{-34}\times3.6\times10^{15}\;\rm J$

$E=2.38\times10^{-18}\;\rm J$

The energy of photon is $2.38\times10^{-18}\;\rm J$ .

For more details about the energy of photon, follow the link given below.

brainly.com/question/15870724.

8 0

2 years ago

Water drips from the nozzle of a shower onto the floor 193 cm below. The drops fall at regular (equal) intervals of time, the fi

Law Incorporation [45]

Answer: 108.81 cm and 48.66 cm

Explanation:

In this, we have to make sure to keep in mind the Gravity effects on the drops. The drops will accelerate when they fall making them travel faster. This means, the velocity is not constant.

What is know:

Height (h) = 193 cm

Gravity (g) = 981 $cm/s^{2}$

Initial Velocity = 0

First, we can know how long it take to the drop to travel to the floor. It can be done with the following equation:

$x = V_{0} t + \frac{1}{2} at^{2}$ (1)

Where:

x is the distance which is 193cm

Vo is the Initial Velocity which is zero

t is the time the time it takes the drop to travel from the shower to the floor

a is the aceleration, which in this case is the gravity.

With the Initial Velocity equals zero the equations simply:

$193 cm = \frac{1}{2}gt^{2}$

To search for the time:

$t =\sqrt{\frac{2*193cm}{981cm/s^{2} } }$

t = 0.627 s

This is the time it takes a drop to fall to the floor, with this time and knowing other 3 drops have driped from the shower by this time. We can calculate how much time it takes the shower to drip each drop.

Time for Drip = t/4

Time for Drip = 0.156

This time is the difference between each drop, using the same equation we can calculate where was each drop, because now it is know how much time had each drop after being drip from the shower.

Our first is already on the floor (193 cm) with 0.627 s, The second drops have been falling for (0.627s - 0.156) 0.471 s and our third drop for (0.627s - 0.156 - 0.156) 0.315 s

We can use (1) to know how far have each drop traveled on these times. We know the Initials Velocity are 0, know we need ot know the distances.

For the second drop:

$x = \frac{1}{2} (981cm/s^{2})(0.471s)^{2}$

$x =108.81 cm$

For the third drop:

$x = \frac{1}{2} (981cm/s^{2})(0.315s)^{2}$

$x = 48.66 cm$

8 0

3 years ago

What fitness level is helpful to know when a person is starting an exercise program

Reil [10]

Answer:

There are various areas that should be considered to access the fitness level of an individual.

Explanation:

Knowing the specific details can help an individual to set realistic fitness goals, monitor their progress, and retain motivation. Fitness is typically measured in four main areas: cardiovascular fitness, muscle strength and endurance, flexibility, and composition of the body. It is important to evaluate these vital components of fitness so that one gets a fair knowledge of present fitness level and plan for the future accordingly.

8 0

4 years ago