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Lady bird [3.3K]
3 years ago
7

Which four equations can be used to solve for acceleration

Physics
1 answer:
emmainna [20.7K]3 years ago
8 0

The four equations for acceleration are obtained from the three equations of motion and from second law of motion.

Explanation:

Acceleration is defined as the rate of change of velocity with respect to time. So the change in velocity with respect to time can be determined using the three equations of motions.

So from the first equation of motion, v = u + at , we can determine the value of acceleration if time taken, final and initial velocity is known. The equation can be re-written as a = \frac{v-u}{t}

Similarly, from the second equation of motion, s = ut + 1/2 at², we can determine the equation for acceleration as a = 2*\frac{s-ut}{t^{2} }

So this is second equation for acceleration.

Then from the third equation of motion, v^{2}- u^{2} = 2* a *s

the acceleration equation is determined as a = \frac{v^{2}-u^{2}  }{2s}

In addition to these three equation, another equation is present to determine the acceleration with respect to force from the Newton's second law of motion. F = Mass × acceleration. From this, acceleration = Force/mass.

So, these are the four equations for acceleration.

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According to the rayleigh criterion, what is the shortest object we could resolve at the 25.0 cm near point with light of wavele
Blizzard [7]

Answer:

6.71 *10^{-5} rad

Explanation:

∅ = \frac{1.22*wavelength}{D} = \frac{1.22*550 * 10^{-9} }{25 * 10^{-2} }

∅ = 6.71 *10^{-5} rad

The minimum resolvable angle = 6.71 *10^{-5} rad

7 0
2 years ago
The sound from a single source can reach point O by two different paths. One path is 20.0 m long and the second path is 21.0 m l
aleksandrvk [35]

Answer:

minimum frequency = 170 Hz

Explanation:

given data

One path long = 20 m

second path long = 21 m

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solution

we get here destructive phase that is path difference of minimum \frac{\lambda}{2}

here  λ is the wavelength of the wave

so path difference will be

21 - 20 = \frac{\lambda}{2}  

λ = 2 m

and

velocity that is express as

velocity = frequency × wavelength    .............1

frequency  = \frac{340}{2}  

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7 0
3 years ago
An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

6 0
3 years ago
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