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3 years ago

To calculate how much rest energy an object contains, you multiply

Physics

1 answer:

Rina8888 [55]3 years ago

5 0

This is what e=mc² means.
So rest energy is RH side of equation that is its mass times speed of light squared.

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Which of these is an appropriate treatment for a deep, bleeding wound?

bija089 [108]

Explanation:

A. Walk the person around

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2 years ago

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A weatherman carried an aneroid barometer from the ground floor to his office atop the Sears Tower in Chicago. On the level grou

Sophie [7]

Answer:

442.36038 m or 1451.31362 ft

Explanation:

$P_1$ = Initial pressure = 30.15 inHg

$P_2$ = Final pressure = 28.607 inHg

$\rho$ = Density of air = 0.075 lb/ft³

$1\ lb/ft^3=16.0185\ kg/m^3$

$1\ in=0.0254\ m$

$1\ m=3.28084\ ft$

Density of mercury = 13560 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Difference in pressure is given by

$P_1-P_2=\rho gh\\\Rightarrow h=\frac{P_1-P_2}{\rho g}\\\Rightarrow h=\frac{(30.15-28.607)\times 13560\times 0.0254\times 9.81}{0.075\times 16.0185\times 9.81}\\\Rightarrow h=442.36038\ m\\\Rightarrow h=442.36038\times 3.28084\\\Rightarrow h=1451.31362\ ft$

The height of the building is 442.36038 m or 1451.31362 ft

4 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

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3 years ago

I’m giving points and an easy answer for y’all hope your having a great day ( and I think some people birthdays are in the 31 so

WARRIOR [948]

Difference between an inherited trait and an acquired trait is that an inherited trait is a trait inherited from a family, such as your eye color or hair color, while a acquired trait is something that you learned yourself such as learning how to read.

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3 years ago

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When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

Veseljchak [2.6K]

Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant

The work done in stretching the rubber band from x = 0 to x = L is
$W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}$

Answer: $\frac{aL^{2}}{2}$

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3 years ago