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Ad libitum [116K]
3 years ago
9

. What is the single most important equation in all of physics?​

Physics
1 answer:
vivado [14]3 years ago
7 0

Answer: F = ma,

Explanation:

the most famous equation in physics, establishing an equivalence between energy and mass. But is this the most important equation in physics? Knowledgeable scientists will tell you no. The most important equation in physics is F = ma, also known as Newton's second law of mechanics.

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DETERMINE THE BCD EQUIVALENT OF 1100111110101001
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Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.

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What is the difference between mechanical digestion and chemical digestion in the mouth
Phoenix [80]
Mechanical digestion is chewing, and chemical digestion is the saliva in your mouth breaking down food.

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3 years ago
Read 2 more answers
The fixed hydraulic cylinder C imparts a constant upward velocity v = 2.2 m/s to the collar B, which slips freely on rod OA. Det
Olenka [21]

Answer:

so angular velocity is 7.13128 sec−1

Explanation:

velocity v = 2.2 m/s

displacement s = 220 mm = 0.220 m

distance d = 510 mm = 0.510 m

to find out

angular velocity

solution

we know that

angular velocity will be velocity ( v)  / (displacement²  +  distance²)   .....1

now put all these value in equation 1 and we get angular velocity i.e.

angular velocity =  velocity ( v)  / (displacement²  +  distance²)

angular velocity = 2.2  / (0.22²  +  0.51²)

angular velocity = 2.2 / 0.3085

angular velocity = 7.13128

so angular velocity is 7.13128 sec−1

6 0
4 years ago
As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equat
DaniilM [7]

Answer:

the volume in terms of time t ⇒ \\V = e^{0.05t} + 399\\

Explanation:

Given that :

\frac{dv}{dt}= kv\\\\\int\limits \, \frac{dv}{v}=  \int\limits \, kdt\\In v = kt + C_1\\v = e^{kt} + C\\400 = e^{k*0} + C\\400 = 1 + C\\C = 400 -1\\C = 399

V = e^{kt} + 399

When v = 300 ;  \frac{dv}{dt}= - 15

then

\frac{dv}{dt}= kv\\\\-15 = 300*k\\\\k = \frac{-15}{300}\\\\\\k = \frac{-1}{20}\\\\k = -0.05

∴ \\V = e^{0.05t} + 399\\

Therefore, the volume in terms of time t ⇒ \\V = e^{0.05t} + 399\\

6 0
3 years ago
A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

8 0
3 years ago
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