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3 years ago

Summarize renewable energy saves water and creates jobs

1 answer:

4 0

Answer: Solar Energy has been helping the supplying companies to be able to hire more people for jobs which helps 12 times more easier than the US, Things like solar, geothermal, and wind can help reduce the water resources which helps to save our water conservation rate which was getting increasingly High.

<em>I hope this explanation had helped you today!</em>

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A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum

Fed [463]

Answer:

983.400345675 hits per second

Explanation:

Radius = 14.2 cm

Record turn rate = 33 rev/min

Bump separation = 0.499 mm

Circumference of the record = $2\pi 0.142=0.89221231362\ m$

Number of bumps in the groove = $\dfrac{0.89221231362}{0.499\times 10^{-3}}=1788.0006285\ bumps$

The rate which the bumps hit the stylus = $33\times\dfrac{1788.0006285}{60}=983.400345675$

The rate at which the bumps hit the stylus 983.400345675 hits per second

8 0

3 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago

Derive the following equations of motion

xz_007 [3.2K]

Answer:

___________________________________

<h3>a. Let us assume a body has initial velocity 'u' and it is subjected to a uniform acceleration 'a' so that the final velocity 'v' after a time interval 't'. Now, By the definition of acceleration, we have:</h3>

$a = \frac{v - u}{t} \\ or \: at = v - u \\ v = u + at \:$

It is first equation of motion.

___________________________________

<h3>b. Let us assume a body moving with an initial velocity 'u'. Let it's final body 'v' after a time interval 't' and the distance travelled by the body becomes 's' then we already have,</h3>

$v = u + at...........(i) \\ s = \frac{u + v}{2} \times t.........(ii)$

Putting the value of v from the equation (i) in equation (ii), we have,

$s= \frac{u + (u + at)}{2} \times t \: \: \\ or \: s = \frac{(2u + at)t}{2} \\ or \: s = \frac{2ut + a {t}^{2} }{2} \\ s = ut + \frac{1}{2} a {t}^{2}$

It is third equation of motion.

________________________________

<h3>c. Let us assume a body moving with an initial velocity 'u'. Let it's final velocity be 'v' after a time and the distance travelled by the body be 's'. We already have,</h3>

$v = u + at.....(i) \\ s = \frac{u + v}{2} \times t......(ii) \\$

$v = u + at \\ or \: at = v - u \\ t = \frac{v - u}{a}$

Putting the value of t from (i) in the equation (ii)

$s = \frac{u + v}{2} \times \frac{v - u}{a} \\ or \: s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ or \: 2as = {v}^{2} - {u}^{2} \\ {v}^{2} = {u}^{2} + 2as$

It is forth equation of motion.

________________________________

Hope this helps...

Good luck on your assignment..

3 0

3 years ago

The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a

SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm= $1900\times 10^{-9}m$

Where $1nm=10^{-9} m$

$\lambda_2=400nm=400\times10^{-9}m$

$\lambda_1=700nm=700\times 10^{-9}m$

We have to find the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

$\theta=sin^{-1}(\frac{m\lambda}{d})$

Using the formula

m=1

$\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})$

$\theta=21.62^{\circ}$

Now, m=2

$\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})$

$\theta_2=24.90^{\circ}$

$\Delta \theta=\theta_2-\theta_1$

$\Delta \theta=24.90-21.62$

$\Delta \theta=3.28^{\circ}$

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

6 0

3 years ago

Recall the raisin cake model of the universe. Our universe is expanding between the galaxies. You measure the recession velocity

Rina8888 [55]

Explanation:

Recession velocity of a galaxy is related to the distance at which the galaxy is located. This relationship is given by the hubble constant, as follows:

$v_r=HD$

Hubble constant is aproximate $70\frac{km/s}{Mpc}$ and 1 megaparsec (Mpc) is 3.26*10^6 light years. Rewriting for D:

$D=\frac{v_r}{H}$

For galaxy A:

$D=\frac{2000\frac{km}{s}}{70\frac{km/s}{Mpc}}\\\\D=28.57Mpc\\\\D=28.57Mpc*\frac{3.26*10^6ly}{1Mpc}=9.31*10^7ly$

For galaxy B:

$D=\frac{6000\frac{km}{s}}{70\frac{km/s}{Mpc}}\\\\D=85.71Mpc\\\\D=85.71Mpc*\frac{3.26*10^6ly}{1Mpc}=2.79*10^8ly$

5 0

3 years ago