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3 years ago

You can hide a row or a column by using the Hide command or by setting the row height or column width to ____________.

Physics

1 answer:

Marianna [84]3 years ago

3 0

Answer:

Explanation:

Suppose this is for Excel. You can hide a row or a column by right-clicking on that row/column header and select row height or column width. After that simply input in 0 and click OK. The row or column will be hidden in the end, just like using a Hide command.

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mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi

valina [46]

Answer:

The answer is

" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

$W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\$

The source of the hooks law is stable,

$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

Assume that solution equation,

$x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\$

These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

Going to compare cos3 t and sin 3 t coefficients from both sides,

The cost3 t is 3A + 3B= 20 coefficients

The sin 3 t is 3B -3A = 0 coefficient

The two equations solved:

$3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\$

Replace the very first equation with the meaning,

$3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\$

equation is

$x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)$

The ultimate plan for both the equation is therefore

$x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)$

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

$x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\$

$x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\$

$c_2=\frac{-64\sqrt{47}}{141}$

$x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$

5 0

3 years ago

Random errors can be reduced by taking repeated measurements.Error and uncertainty are interchangeable words that describe the s

fiasKO [112]

repeated mesurement can reduce the error

it is true

if you take any mesurement repeatedly and the average is taken, the error will be less

5 0

1 year ago

Which considerations are used to calculate a windchill factor? Select two options.

vlada-n [284]

Wind speed and air temperature are used to calculate a windchill factor.

<u>Explanation:</u>

<u></u>

Wind-chill factor is the reduction of body temperature due to the passing flow of lower-temperature air.

The air temperature value is always higher than the wind chill numbers. the heat index will be used if the apparent temperature is higher than the air temperature.So, Wind speed and air temperature are mainly used to calculate a windchill factor.

There are many ways, the surface loses its heat through conduction, evaporation,radiation, and convection.The rate of convection depends on the difference in temperature between the surface and the fluid surrounding the surface and the velocity of that fluid with respect to the surface. The air around the warm surface will be heated, an insulating layer of warm air forms against the surface.The layer becomes a boundary between two. As the wind speed is high the surface cools down rapidly.

7 0

3 years ago

Read 2 more answers

(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

PilotLPTM [1.2K]

Answer:

Part a)

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

$v = 2.18 \times 10^5 m/s$

Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

$T = 8.2 \times 10^{15} s$

Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

5 0

4 years ago

PLEASE HELP ASAP!! What are 3 types of forces acting on your marshmallow (or other flying object) when it impacts it’s target? C

Alekssandra [29.7K]

•Every action has an equal and opposite reaction (the object is putting force on the target, and the target is putting an equal amount of force back)
•Am object in motion (the object) will stay in motion unless an outside force acts upon it (the Target)

And as for the third one I really don’t know, those seem to be the only two, I’m sorry. I did what a could, Hope it kinda helps :)

8 0

4 years ago