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3 years ago

10

What angular velocity in revolution per second is needed for a centrifuge to produce an centripetal acceleration of a=9800m/s sq

uare at radius 10 cm.

1 answer:

max2010maxim [7]3 years ago

7 0

Option (d) is correct.

Explanation:

centripetal acceleration = a = r ω²

r= radius= 10 cm= 0.1 m

ω= angular velocity

9800= (0.1) ω²

ω=313 rad/sec

ω= 313 rad/s * (1 rev/2π rad)

ω=49.8 rev/s

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One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s

DedPeter [7]

Complete Question

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cm across, and you estimate that the distance from the window shade to the wall is about 5 m.

Required:

Estimate the diameter of the pinhole.

Answer:

The diameter is $d =0.000336 m$

Explanation:

From the question we are told that

The central maxima is $D= 2cm = \frac{2}{100} = 0.02m$

The distance from the window shade is $L = 5m$

The average wavelength of the sun is mathematically evaluated as

$\lambda_{ave } = \frac{\lambda_i + \lambda_f}{2}$

Generally the visible light spectrum has a wavelength range between 400 nm to 700 nm

So the initial wavelength of the sun is $\lambda _i = 400nm$

and the final wavelength is $\lambda_f = 700nm$

Substituting this into the above equation

$\lambda_{sun} = \frac{400nm +700nm}{2}$

$= 550nm$

The diameter is evaluated as

$d = \frac{2.44 \lambda_{sun} L}{D}$

substituting values

$d = \frac{2.44 * 550*10^{-9} * 5 }{0.02}$

$d =0.000336 m$

5 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago

Chanice drives her scooter 7 kilometers north. she stops for lunch and then drives 5 kilometers and then 1 km east again. what d

lord [1]

Distance covered is given as follows

1). 7 km North

2). 5 km North

3). 1 km East

Now total distance covered will be given as

$d = d_1 + d_2 + d_3$

$d = 7 km + 5 km + 1 km$

$d = 13 km$

Now in order to find the displacement we will show all with their directions

$\vec d_1 = 7 + 5 = 12 km$ towards North

$\vec d_2 = 1 km$ towards East

So total displacement is

$\vec d = \vec d_1 + \vec d_2$

$\vec d = 12 \hat j + 1 \hat i$

so net displacement will be

$d = \sqrt{12^2 + 1^2} = 12.04 km$

so displacement is 12.04 km

6 0

4 years ago

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f

tatiyna

Answer:

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

$Em_{f}$ = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m $v_{cm }^{2}$ + ½ $I_{cm}$ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

4 0

3 years ago

Calculate the Poynting vector at the surface of the filament, associated with the static electric field producing the current an

Vesnalui [34]

We anticipate a constant Poynting vector of magnitude since the hot resistor will be emitting heat and none of the electric or magnetic fields will change over time.

S = P/A

= I2R/ 2πrL

= 332 kW/m2

Always pointing away from the wire, this Poynting vector.

<h3>What is the Poynting vector?</h3>

Describes the size and direction of the energy flow in electromagnetic waves using a Poynting vector. It bears the name of the 1884 invention of English physicist John Henry Poynting. It stands for the electromagnetic field's directional energy flux or power flow. The Poynting vector is significant in a static electromagnetic field because it determines the direction of energy flow in an electromagnetic field. This vector represents the radiation pressure of an electromagnetic wave and points in its direction of propagation.

To learn more about Poynting vector, visit:

<u>brainly.com/question/17330899</u>

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1 year ago