What is the focal length of a lens

For elliptical obits: the direction of the velocity of the satellite is _______________________ (always, seldom, never) perpendi

alexandr1967 [171]

Answer:

For elliptical orbits: seldom

For circular orbits: always

Explanation:

We start by analzying a circular orbit.

For an object moving in circular orbit, the direction of the acceleration (centripetal acceleration) is always perpendicular to the direction of motion of the object.

Since acceleration has the same direction of the force (according to Newton's second law of motion), this means that the direction of the force (the centripetal force) is always perpendicular to the velocity of the object.

So for a circular orbit,

the direction of the velocity of the satellite is always perpendicular to the net force acting upon the satellite.

Now we analyze an elliptical orbit.

An elliptical orbit correponds to a circular orbit "stretched". This means that there are only 4 points along the orbit in which the acceleration (and therefore, the net force) is perpendicular to the direction of motion (and so, to the velocity) of the satellite. These points are the 4 points corresponding to the intersections between the axes of the ellipse and the orbit itself.

Therefore, for an elliptical orbit,

the direction of the velocity of the satellite is seldom perpendicular to the net force acting upon the satellite.

7 0

3 years ago

23. A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exer

Sophie [7]

Answer:

Explanation:

Given

mass of child=22 kg

weight of child $=mg=22\times 9.8=215.6 N$

$N=40 rev/min$

$\omega _1=\frac{2\pi 40}{60}=4.189 rad/s$

distance from, center $r_1=1.25 m$

Centripetal Force she exerts to stay on

$F_1=m(\omega _1)^2\times r_1$

$F_1=22\times (4.189)^2\times 1.25$

$F_1=482.56 N$

$F_1 is 2.23$ times of child weight

Now if

$N_2=3 rev/min$

$\omega _2=0.3142 rad/s$

$r_2=8 m$

$F_2=m(\omega _2)^2\times r_2$

$F_2=22\times (0.3142)^2\times 8=17.37 N$

$F_2\ is\ 0.08$ times of child weight

6 0

3 years ago

A photoelectric-effect experiment finds a stopping potentialof 1.93V when light of 200nm is used to illuminate thecathode.

Readme [11.4K]

Answer:

a) Tantalum

b) 1.93 V

Explanation:

The energy of the incident photon= hc/λ

h= Plank's constant=6.63×10^-34 Is

c= speed of light = 3×10^8 ms-1

λ= wavelength of incident photon

E= 6.63×10^-34 × 3×10^8/ 200×10^-9

E= 0.099×10^-17

E= 9.9×10^-19 J

The kinetic energy of the electron = eV

Where;

e= electronic charge = 1.6×10^-19 C

V= 1.93 V

KE= 1.6×10^-19 C × 1.93 V

KE= 3.1 ×10^-19 J

From Einstein's photoelectric equation;

KE= E -Wo

Wo= E -KE

Wo=9.9×10^-19 J - 3.1 ×10^-19 J

Wo= 6.8×10^-19 J

Wo= 6.8×10^-19 J/1.6×10^-19

Wo= 4.25 ev

The metal is Tantalum

b) the stopping potential remains 1.93 V because intensity of incident photon has no effect on the stopping potential.

4 0

3 years ago

Tires are rotated to A. preserve the warranty. B. preserve the front-wheel alignment. C. prevent tire scrubbing. D. promote even

andre [41]

The answer is B preserve the front wheel alignment.

3 0

3 years ago

The linear charge density on the inner conductor is and the linear charge density on the outer conductor is

Lubov Fominskaja [6]

Complete Question

The complete question is shown on the first uploaded image (reference for Photobucket )

Answer:

The electric field is $E = -1.3 *10^{-4} \ N/C$

Explanation:

From the question we are told that

The linear charge density on the inner conductor is $\lambda _i = -26.8 nC/m = -26.8 *10^{-9} C/m$

The linear charge density on the outer conductor is

$\lambda_o = -60.0 nC/m = -60.0 *10^{-9} \ C/m$

The position of interest is r = 37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

$E (2 \pi r l) = \frac{ \lambda_i }{\epsilon_o}$

=> $E = \frac{\lambda _i }{2 \pi * \epsilon_o * r}$

substituting values

$E = \frac{ -26 *10^{-9} }{2 * 3.142 * 8.85 *10^{-12} * 0.0373}$

$E = -1.3 *10^{-4} \ N/C$

The negative sign tell us that the direction of the electric field is radially inwards

=> $|E| = 1.3 *10^{-4} \ N/C$

5 0

4 years ago