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Harlamova29_29 [7]
3 years ago
9

A 2-inch, f/6 telescope has a 3-inch eyepiece focal. Its magnifying power is:

Physics
1 answer:
snow_tiger [21]3 years ago
5 0
The choices can be found elsewhere and as follows:

a. 5X 
b. 6X 
c. 15X 
<span>d. 60X 
</span>
I believe the correct answer from the choices listed is option C. <span>A 2-inch, f/6 telescope has a 3-inch eyepiece focal. Its magnifying power is 15x. Hope this answers the question. Have a nice day.</span>
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A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed
Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

\int EdS=\frac{Q_{in}}{\epsilon_o}   (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

\int EdS=ES=E(4\pi r^2)   (2)

Qin is calculate by using the charge density:

Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho  (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}

Next, you use the results of (3), (2) and (1):

E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}

hence, the electric field is 1580594.95 N/C

7 0
3 years ago
PLEASE HELP!!!! Explain why an object made from aluminum will not stick to a magnet​
uranmaximum [27]
Because aluminum is not made out of the same material as a magnet would be
6 0
3 years ago
A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformati
Slav-nsk [51]

The question is incomplete. The complete question is :

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:

J(t) = Jo (1 - exp (-t/to))

Where,

Jo= 3m^2/ GPA

to= 200s

Determine

a) the strain after 100's (before stress is reversed)

b) the residual strain when stress falls to zero.

Answer:

a)-60GPA

b) 0

Explanation:

Given t= 0,

σ = 20Mpa

Change in σ= 0.2Mpas^-1

For creep compliance material,

J(t) = Jo (1 - exp (-t/to))

J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa

a) t= 100s

E(t)= ΔσJ (t - Jo)

= 0.2 × 3 ( 100 - 200 )

= 0.6 (-100)

= - 60 GPA

Residual strain, σ= 0

E(t)= Jσ (Jo) ∫t (t - Jo) dt

3 × 0 × 200 ∫t (t - Jo) dt

E(t) = 0

5 0
3 years ago
Write down Newton's second law in terms of momentum and acceleration. Write down this law in the form of differential equation (
svetoff [14.1K]

Explanation:

According to Newton's second law of motion, the rate of change of momentum is directly proportional to the applied unbalanced force. The mathematical expression is given by:

F=\dfrac{d(mv)}{dt}

Where

F is the applied force

m is the mass of the object

v is the velocity with which it is moving

F=m\dfrac{dv}{dt}

Momentum of a particle is given by the product of mass and velocity as :

p=mv

Hence, this is the required solution.

3 0
3 years ago
Which scientific law describes the relationship between action and reaction force pairs? Newton's third law of motion Kepler's l
Hitman42 [59]

Newton's third law of motion

Explanation:

Newton's third law of motion states that:

<em>"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force (reaction force) on object A"</em>

It is important to note that this law is always valid, even when it seems it is not.

Consider for example the gravitational force that the Earth exerts on your body (= your weight). We can say that this is the action force. It may seems that there is no reaction force in this case. However, this is not true: in fact, your body also exerts an equal and opposite force on the Earth, and this is the reaction force. The reason that explains why we don't notice any effect on Earth due to this force is that the mass of the Earth is much larger than your mass, therefore the acceleration produced on the Earth because of the force you apply is negligible.

It is also important to note that the action-reaction pair of forces always act on two different objects, so they never appear in the same free-body diagram.

Learn more about Newton's third law of motion:

brainly.com/question/11411375

#LearnwithBrainly

7 0
3 years ago
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