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3 years ago

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In an experiment, an object is released from rest from the top of a building. Its speed is measured as it reaches a point that i

s a distance d from the point of release. If this distance was doubled, what would the new speed be, assuming air resistance is negligible?

1 answer:

NikAS [45]3 years ago

4 0

Answer:

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

Explanation:

As we know that the air friction or resistance due to air is neglected then we can use the equation of kinematics here

$v_f^2 - v_i^2 = 2 a d$

since we released it from rest so we have

$v_i = 0$

so here we have

$v_f = \sqrt{2gd}$

now if the distance is double then we have

$v_f' = \sqrt{2g(2d)}$

now from above two equations we can say that

$v_f' = \sqrt2 v_f$

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

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A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t

Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

3 0

2 years ago

What can you infer about a wave with a short wavelength?

raketka [301]

Answer:

- It can be infer that it has a lower frequency.

<em>In the case of electromagnetic waves.</em>

- A short wavelength means a lower energy,

Explanation:

The wavelength is the distance between two consecutive crests or valleys while the frequency is the number of crests that pass for a specific point in an interval of time.

For example, a person makes laundry once a weak.

In this example, the event is represented by the laundry and the interval of time is once a weak

The velocity of a wave is defined as:

$v = \nu \cdot \lambda$ (1)

Where $nu$ is the frequency and $\lambda$ is the wavelenth

$\lambda = \frac{v}{\nu}$ (2)

Notice from equation 2 that the wavelength is inversely proportional to the frequency (when the wavelength increases the frequency decreases).

In the case of electromagnetic waves, a short wavelength means a lower energy, as it can be seen in equation 4 (inversely proportional).

$E = h\nu$ (3)

$E = \frac{hc}{\lambda}$ (4)

8 0

2 years ago

Read 2 more answers

Is the energy of a wave is calculated by squaring the frequency of the wave?

Misha Larkins [42]

Nope.
Energy is directly proportional to frequency. and when you calculate energy, you multiply frequency with a constant number called "Planck's Constant"

E = hf

Hope this helps!

5 0

3 years ago

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0

3 years ago

The true brightness of an object is called its

g100num [7]

<span>The true brightness of an object is called its luminosity. It is the total amount of energy emitted by bright or meteorological objects over a period of time. It has the SI unit of joules per second or watts. So the answer is letter A. Intensity is the measure of how strong the substance or object is when it projects something. Magnitude is a measure of how great is the size the object produces. Viscosity is the measure of flow of a substance.</span>

8 0

3 years ago

Read 2 more answers