List these things in order of size, starting with the smallest thing:

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

3 0

3 years ago

The visible spectrum refers to the portion of the electromagnetic spectrum that we ________.

yKpoI14uk [10]

<h2>Answer: can see</h2>

Explanation:

The portion visible by the human eye of the electromagnetic spectrum is between 380 nm (violet-blue) and 780 nm (red) approximately. Which means this part of the spectrum is located between ultraviolet light and infrared light.

Note the fact only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

Therefore:

<h2>The visible spectrum refers to the portion of the electromagnetic spectrum that <u>we </u><u>can see</u></h2>

8 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 8.0 s, it rotates 35 rad.

Harman [31]

Answer:

(a) 1.093 rad/s^2

(b) 4.375 rad/s

(c) 8.744 rad/s

(d) 67.845 rad

Explanation:

initial angular velocity, ωo = 0

time, t = 8s

angular displacement, θ = 35 rad

(a) Let α be the angular acceleration.

Use second equation of motion for rotational motion

$\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}$

By substituting the values

35 = 0 + 0.5 x α x 8 x 8

α = 1.093 rad/s^2

(b) The average angular velocity is defined as the ratio of total angular displacement to the total time taken .

Average angular velocity = 35 / 8 = 4.375 rad/s

(c) Let ω be the instantaneous angular velocity at t = 8 s

Use first equation of motion for rotational motion

ω = ωo + αt

ω = 0 + 1.093 x 8 = 8.744 rad/s

(d) Let in next 5 seconds the angular displacement is θ.

$\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}$

By substituting the values

θ = 8.744 x 5 + 0.5 x 1.093 x 5 x 5

θ = 67.845 rad

8 0

4 years ago

A wheel decelerates from 13.5 rad s−1 to 6.0 rad s−1 in 7 s. Calculate the angular displacement

Igoryamba

Answer:

<em>Angular displacement=68.25 rad</em>

Explanation:

<u>Circular Motion</u>

If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

$\displaystyle \alpha=\frac{\omega_f-\omega_o}{t}$

The angular displacement is given by:

$\displaystyle \theta=\omega_o.t+\frac{\alpha.t^2}{2}$

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:

$\displaystyle \alpha=\frac{6-13.5}{7}$

$\displaystyle \alpha=\frac{-7.5}{7}$

$\displaystyle \alpha=-1.071 \ rad/s^2$

Thus, the angular displacement is:

$\displaystyle \theta=13.5*7+\frac{-1.071*7^2}{2}$

$\displaystyle \theta=94.5-26.25$

$\boxed{\displaystyle \theta=68.25\ rad}$

Angular displacement=68.25 rad

3 0

3 years ago

Light passes through a pair of very thin parallel slits. The resulting interference pattern is viewed far from the slits at vari

butalik [34]

Answer:

Intensity of the next bright fringe will remain same.

Explanation:

The question is based on Young's double slit experiment, since its about bright fringe, the interference here is constructive.

Young's condition for constructive interference is given by:

dsin $\theta = n\lambda$

where,

d = slits distance from eachother or width of the slits

$\lambda$ = wavelength

n = interferance order

Also, we know that in Young's experiment, the fringe intensity is given by:

$I' = 4Icos^{2}\phi$

where,

$\phi$ = phase difference

Therefore, in absence of phase difference i.e., $\phi = 0$ , the intensity of the next bright fringe will not change and it will remain same.

4 0

3 years ago