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sladkih [1.3K]
3 years ago
6

Which statement is the best interpretation of the ray diagram shown?

Physics
1 answer:
n200080 [17]3 years ago
6 0

Answer:

Explanation:

First off, this lens is concave. Second, the image is obviously smaller, and third, the only thing that is NOT obvious, is the fact that real images are always upside down and virtual images are always right-side-up. So the choice you're looking for is D.

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How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a po
miss Akunina [59]

Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.

An electron is released from rest at a distance of 0 m  from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² .  How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?

Answer:

Work=1.06×10⁻²¹J

Explanation:

Given Data

Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²

Charge density σ=3.90×10⁻¹² C/m²

The electron moves a distance d=3.00×10⁻²m

Electron charge e=-1.6×10⁻¹⁹C

To find

Work done

Solution

The electric field due is sheet is given as

E=σ/2ε₀

E=\frac{3.90*10^{-12}C/m^{2}  }{2(8.85*10^{-12}C^{2} /N.m^{2} )}\\ E=0.22V/m

Now we need to find force on electron

F=eE\\F=(1.6*10^{-19}C )(0.22V/m)\\F=3.525*10^{-20}N

Now for Work done on the electron

W=F*d\\W=(3.525*10^{-20} N)(3.00*10^{-2}m)\\W=1.06*10^{-21}J

4 0
4 years ago
You are riding on a carousel that is rotating at a constant 24 rpm. It has an inside radius of 4 ftand outside radius of 12 ft.
Nookie1986 [14]

Answer:

magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

Explanation:

Given the data in the question;

Speed of carousel N = 24 rpm

From the diagram below, selected path direction defines the Axis of slip.

Hence, The Coriolis is acting along the axis of transmission

Now, we determine the angular speed ω of the carousel.

ω = 2πN / 60

we substitute in the value of N

ω = (2π × 24) / 60

ω = 2.5133 rad/s

Next, we convert the given velocity from mph to ft/s

we know that; 1 mph = 1.4667 ft/s

so

V_{slip = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s

Now, we determine the magnitude of the Coriolis acceleration

a_c = 2( V_{slip × ω )

we substitute

a_c = 2( 8.8002 ft/s × 2.5133 rad/s )

a_c = 44.235 ft/s²

Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

7 0
3 years ago
a bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s² for 10s. Find the distance moved by it during this interval o
BaLLatris [955]

Answer:

90 m

Explanation:

Acceleration, a=\frac {v-u}{t} where v and u are final and initial velocities respectively, t is the time taken

Substituting 1 m/s^{2} for a,  4 m/s for u and 10 s for t then

1*10=v-4

v=14 m/s

From kinematic equations

v^{2}=u^{2}+2as

Making s the subject then

s=\frac {v^{2}-u^{2}}{2a}=\frac {14^{2}-4^{2}}{2\times 1}=90 m

6 0
3 years ago
The cycling of the air is a natural phenomenon as shown above is the result of
Ket [755]

Answer:

natural convection

............

6 0
3 years ago
Read 2 more answers
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
4 years ago
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