Ask question

Ask question

All categories

andrew-mc [135]

4 years ago

6

A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the

average speed of the blade tip in the helicopter’s frame of reference. (b) What is its average velocity over one revolution?\

1 answer:

professor190 [17]4 years ago

7 0

Explanation:

It is given that,

A helicopter blade spins at exactly 100 revolutions per minute.

Its tip is 5.00 m from the center of rotation, r = 5 m

(a) Let v is the average speed of the blade tip in the helicopter’s frame of reference. Distance covered by the helicopter, $d=2\pi r$

In 100 revolutions, $d=200\pi r=1000\pi$

So, average speed of the blade tip in one second is given by :

$v=\dfrac{d}{t}$

$v=\dfrac{1000\pi\ m}{60\ s}$

v = 52.35 m/s

(b) The average velocity over one revolution is zero because the net displacement in one rotation is 0.

Hence, this is the required solution.

You might be interested in

PLZ HELP!!WILL MARK THE BRAINLIEST!!The diagram shows a chromosome pair for an offspring.Which best describes the inheritance of

Dmitrij [34]

Answer:

The offspring is homozygous for eye color

Explanation: I had this question on my test and this was the answer, i hope this helps :)

6 0

3 years ago

Read 2 more answers

Where are all stars bornWhat is the brightest star in the night sky? What is its magnitude?

sergij07 [2.7K]

Answer:Stars are born within the clouds of dust and scattered throughout most galaxies.It's the star Sirius in the constellation Canis Major, brightest star in the sky. The bright planet Venus is also up before dawn now. But you'll know Sirius, because Orion's Belt always points to it.With an apparent magnitude of −1.46

Explanation: Brainliest Please!!!!

5 0

3 years ago

A 1.75 kg stone falls from the top of a cliff and strikes the ground at a speed of 58.8 m/s. What is the height of the cliff??

Reil [10]

Answer:

172.9m

Explanation:

h = 1/2 gt^

First calculat for t using

v = gt

t = v/g = 58.8/10

= 5.88secs

now h = 1/2 x 10 x 5.88^2

h =1/2 x 10 x 34.57

= 345.74/2

= 172.9m

6 0

3 years ago

Why is the sign of the emf of the second peak opposite to the sign of the first peak?

eimsori [14]

Because AC emf is a sine wave

7 0

3 years ago

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago