Fomula for atmospheric presure

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0

1 year ago

A truck going 15 kmôŹ°€h has a head-on collision with a small car going 30 kmôŹ°€h. Which statement best describes the situation

zlopas [31]

1. e) None of the above is necessarily true.

2.d) Without knowing the mass of the boat and the sack, we cannot tell.

5 0

3 years ago

A fireman of 80 kg slides down a pole.When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s . By how

levacccp [35]

Answer:

3099 J

Explanation:

The increase in thermal energy corresponds to the mechanical energy lost in the process.

The mechanical energy is given by the sum of gravitational potential energy and kinetic energy of the fireman:

$E=K+U$

At the top of the pole, the fireman has no kinetic energy, so all his mechanical energy is just potential energy:

$E=U=mgh=(80 kg)(9.8 m/s^2)(4.2 m)=3293 J$

When the fireman reaches the bottom, he has no gravitational potential energy, so his mechanical energy is just given by his kinetic energy:

$E=K=\frac{1}{2}mv^2=\frac{1}{2}(80 kg)(2.2 m/s)^2=194 J$

So, the loss in mechanical energy was

$\Delta E=3293 J-194 J=3099 J$

and this corresponds to the increase in thermal energy.

7 0

3 years ago

The deviation of a value (x) from the mean of the sample divided by the standard deviation (s)of the sample is known as the "z"

Ugo [173]

Answer:

true.

Explanation:

z-score gives an idea of how far the value of data is from the mean point. Basically is the measure of a standard deviation below or above the mean.

More the deviation of the data from the mean more is the standard deviation.

Z-score or value is calculated by the expression given below:

$z = \dfrac{x-\mu}{\sigma}$

x is the value

μ is the mean

σ is the standard deviation

Hence, the statement given is true.

6 0

3 years ago

A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo

NISA [10]

Explanation:

The given data is as follows.

Mass of small bucket (m) = 4 kg

Mass of big bucket (M) = 12 kg

Initial velocity ( $v_{o}$ ) = 0 m/s

Final velocity ( $v_{f}$ ) = ?

Height $H_{o} = h_{f}$ = 2 m

and, $H_{f} = h_{o}$ = 0 m

Now, according to the law of conservation of energy

starting conditions = final conditions

$\frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}$

$\frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)$

235.44 = $8V^{2}_{f}$ + 78.48

$V_{f}$ = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

3 0

3 years ago

Fomula for atmospheric presure​

Fomula for atmospheric presure