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3 years ago

Which calculation is an example of velocity

Physics

2 answers:

stellarik [79]3 years ago

8 0

Answer: 10 m/s to the right

Explanation: I did the test

Dmitry_Shevchenko [17]3 years ago

3 0

Answer:

Yes

Explanation:

sun rays towards the earth is an example of velocity

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A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

3 years ago

If you double the distance between you and the center of Earth, what happens to the strength of the gravitational field you expe

Lostsunrise [7]

Answer:

The strength of gravity decreases.

An example of that would be if you were in space; you float around because there's no gravity.

7 0

3 years ago

Which diagram is the best model for a solid?<br> Substance A<br> Substance B<br> О Substance C

mina [271]

Answer:

This link was diagram

Explanation:

https://doubtnut.app.link/FnsNC80Dccb

7 0

3 years ago

Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line

Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

$y=\frac{m\lambda D}{d}$

where

m is the order of the maximum

$\lambda$ is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

$\lambda=632.8 nm = 632.8\cdot 10^{-9} m$ is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

$n=520 lines/mm$ is the number of lines per mm, so the spacing between two lines is

$d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m$

Therefore, substituting m = 1, we find:

$y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m$

So, on the distant screen, there is 1 bright fringe every 33 cm.

6 0

3 years ago

An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. if the angular ac

elena-s [515]

First of all, we need to convert the angular speed from rev/min into rev/s:

$\omega_f=300 rev/min=5 rev/s$

The angular acceleration is the variation of angular speed divided by the time:

$\alpha=\frac{\omega_f-\omega_i}{t}=\frac{5 rev/s-0}{2 s}=2.5 rev/s^2$

And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

$\theta(t)=\frac{1}{2}\alpha t^2 =\frac{1}{2}(2.5 rev/s^2)(2 s)^2=5 rev$

so, 5 revolutions.

3 0

3 years ago