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3 years ago

Look at the graph

Physics

2 answers:

dmitriy555 [2]3 years ago

7 0

Answer:

Option (C)

Explanation:

The slope of the lie is given by the ratio of Y axis intercept to the X axis intercept.

It is also defined as

TanФ = ΔY / ΔX

Here, ΔY = 600 g (nearly)

ΔX = 70 cm^3

So, the slope is 8.57 g/cm^3

The slope of mass Vs volume gives the density.

So, the density is 8.57 g/cm^3.

viva [34]3 years ago

4 0

Here, Carefully look at the graph.
When it is on x=10, it is approximately 10, (slightly less than 10)
Closest value would be 90, so y/x = 90/10 = 9

So, the density of the graph would be 9 g/cm³

In short, Your Answer would be Option D

Hope this helps!

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Two waves are superposed. one wave has an amplitude of 5 cm, and other has an amplitude of 4 cm. What is the resultant amplitude

Vsevolod [243]

If both waves have the same wavelength, then the amplitude of
their sum could be anything between 1 cm and 9 cm, depending
on the phase angle between them.

If the waves have different wavelengths, then the resultant is a beat
with an amplitude of 9 cm.

4 0

3 years ago

(b) The particle displacement y of air molecules due to a sound wave is given by y 4m and w = = 0.008 cos wt sin kz. Where k - m

serious [3.7K]

The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.

<h3>What's the distance between consecutive nodes of the displacement of air molecules?</h3>

Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.
So, distance between consecutive nodes = wavelength = 2π÷k

= 2π/(4π÷m)

= m/2

<h3>What's the amplitude after 0.56s of the displacement of air molecules?</h3>

Displacement after 0.56 s = 0.008×cos(50π×0.56s)

=1.75×10^(-4) m

Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.

Calculate:

I) the distance between 2 consecutive nodes

ii) the amplitude after 0.565s

Learn more about the wavelength here:

brainly.com/question/10750459

#SPJ1

8 0

2 years ago

NEWTONS SECOND LAW LAB REPORT

hram777 [196]

Answer:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Explanation:

here this may help.

8 0

3 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

7 0

3 years ago

Two planets, Dean and Sam, orbit the Sun. They each have with circular orbits, but orbit at different distances from the Sun. De

lyudmila [28]

Answer:

The correct answer is Dean has a period greater than San

Explanation:

Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.

T² = (4π² / G M) r³

When applying this equation to our case, the planet with a greater orbit must have a greater period.

Consequently Dean must have a period greater than San which has the smallest orbit

The correct answer is Dean has a period greater than San

3 0

3 years ago