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3 years ago

5

Which is true concerning the acceleration due to gravity? A. It decreases with increasing altitude. B. It is different for diffe

rent objects in free fall. C. It is a fundamental quantity. D. It is a universal constant. E. all of these

1 answer:

Kisachek [45]3 years ago

5 0

Answer:

The correct answer is option 'a': It decreases with increase in altitude

Explanation:

Acceleration due to gravity is the acceleration that a body is subjected to when it is freely dropped from a height from surface of any planet, ignoring the resistance that the object may face in it's motion such as drag due to any fluid.. The acceleration due to gravity is same for all the objects and is independent of their masses, it only depends on the mass of the planet and the radius of the planet on which the object is dropped. it's values varies with:

1) Depth from surface of planet.

2)Height from surface of planet.

3) Latitude of the object.

Hence it neither is a fundamental quantity nor an universal constant.

The variation of acceleration due to gravity with height can be mathematically written as:

$g(h)=g_{surface}(1-\frac{2h}{R_{planet}})$

where,

R is the radius of the planet

$g_{surface}$ is value of acceleration due to gravity at surface.

hence we can see that upon increase in altitude the value of 'g' goes on decreasing.

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The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper

-BARSIC- [3]

To solve this problem we will apply the normal distribution, with which we will obtain the probability that the given event will occur. Concepts such as the mean and standard deviation will be present throughout the solution of the problem. Increasing or decreasing the average would change the location or center point of the curve. The change in the standard deviation would lead to the change in the dispersion of the data. As the standard deviation increases, the curve would become flatter.

Let X be the output voltage of power supply

X∼N $(5,0.02^2)$

A

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$P(4.95$

$P(4.95$

$P(4.95$

$P(4.95$

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4 years ago

Whats the difference between speed and velocity of an object

ss7ja [257]

The difference between speed and velocity is that the speed is a scalar quantity which means that you can say that this object has a speed of x m/s but you don't have to define its direction
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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

3 years ago

A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N

vaieri [72.5K]

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

$p = \frac{nRT}{V}$

<u>Where:</u>

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

$n = \frac{m}{M}$

<u>Where:</u>

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

$n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles$

Now, the temperature can be found using the following equation:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

<u>Where:</u>

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

$v_{rms}$ : is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

$T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K$

Finally, we can find the pressure of the gas:

$p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm$

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

8 0

3 years ago

Which object is I’m equilibrium?

IrinaK [193]

Equilibrium is a state of no motion. If donuts are on the table, they are in equilibrium. Literally anything that is NOT moving relative to the earth is in equilibrium. The chair is in equilibrium, the table is in equilibrium, the cookies are in equilibrium, and so on.

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3 years ago