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1 year ago

Choose the force diagram (freebody diagram) that best represents the forces acting on the object.

Physics

1 answer:

Levart [38]1 year ago

6 0

Answer:

Diagram B because a plate resting on a table has gravity acting on it and normal/air resistance force working against the gravity with the same amount of force. Forces that are balanced don't move.

Explanation:

I hope this helps

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(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

7 0

3 years ago

Given a block of glass that is semi-circular, a laser pointer, protractor/ruler, and index card, design your own experiment to o

Charra [1.4K]

Answer:

we go up the ramp there is a point where the beam is reflected inside the block, we carefully step back to the point where the beam is horizontal, we measure this angle which is our critical angle.

Explanation:

To design the experiment of measuring the critical angle, we describe the phenomenon, when the light passes from a medium with a higher refractive index to one with a lower index, it separates from the normal one and the Critical Angle is defined as the Angle for which the refraction occurs at 90º

n₂ sin θ₂ = n₁ sin 90

n₁ / n₂ = sin θ₂

As we can see, we have to measure the angle with which the laser touches the exit surface of the glass block.

Design of the experiment:

We place the glass block on the ramp and at the top we hit the conveyor for half the angle, we climb the block on the ramp and see that the angle of incidence of lightning on the exit face changes, part of the beam comes out of the glass , we see it by dispersion in the particles of dirty in the air; Maybe the conveyor or the laser should be moved slightly so that the beam touches the point of origin on the conveyor.

When we go up the ramp there is a point where the beam is reflected inside the block, we carefully step back to the point where the beam is horizontal, we measure this angle which is our critical angle.

5 0

3 years ago

Read 2 more answers

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$