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3 years ago

Refer the attached photo

Physics

1 answer:

Fofino [41]3 years ago

8 0

Answer:

Explanation:

since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet

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From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of

alexandr1967 [171]

Answer:

The value is $h = 11930 \ m$

Explanation:

From the question we are told that

The temperature is $T = 300 \ K$

Generally the root mean square speed of the oxygen molecules is mathematically represented as

$v = \sqrt{\frac{3 * R * T }{M} } = \sqrt{ 2 * g * h }$

Here R is the gas constant with a value $R = 8.314 \ J\cdot K^{-1} \cdot \ mol^{-1}$

M is the molar mass of oxygen molecule with value $M = 0.032 \ kg /mol$

$\frac{3 * 8.314 * 300 }{0.032} = 2 * 9.8 * h$

=> $h = 11930 \ m$

4 0

2 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

7 0

3 years ago

Who needs to know physics when they are making a satellite

vladimir2022 [97]

First satellite was Sputnik and it was a race then to send satellites!! Everyone starts experimenting how to launch.

first try was on a dog , Laika !! It was send to space as it is long after sputnik is placed in orbit !!

Now you know everyone knows rocket science and the most active are ISRO (India) and NASA (USA) !!

4 0

3 years ago

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

2 years ago

Two identical waves are destructively interfere. What will happen to the resulting wave?

storchak [24]

Answer:

Because the disturbances are in opposite directions for this superposition, the resulting amplitude is zero for pure destructive interference

Explanation:

8 0

2 years ago