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zloy xaker [14]
3 years ago
5

A convex lens is placed on a flat glass plate and illuminated from above with monochromatic red light. When viewed from above, c

oncentric bands of red and dark are observed. What does one observe at the exact center of the lens where the lens and the glass plate are in direct contact
Physics
1 answer:
kipiarov [429]3 years ago
4 0

Answer:xit wou,ld becevemtjyrjtdjtdhtdjyrkuf

Explanation:

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A cannon is fired horizontally from atop a 40.0 m tower. The cannonball travels 175 m horizontally before it strikes the ground.
Anastasy [175]

The vertical movement of the projectile is described by:

y = H - gt² / 2

When the cannonball is on the ground, y= 0 so:

0 = H - gt² /2

Solving for t:

t = \sqrt{\dfrac{2H}{g}} = \sqrt{\dfrac{2(40\;m)}{9.8\;m/s^2}} = 2.8\;s

The horizontally movement of the projectile is described by:

x = v₀t

Solving for v₀:

v₀ = x/t

v₀ = 175 m / 2.8 s

v₀ = 61.5 m/s

3 0
3 years ago
when an electric current is flowing through a wire which of the following is also produced energy domain field force
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When an electric current is flowing through a wire, there is a magnetic <em>field</em> produced.  The lines of magnetic force are circles around the wire.

4 0
3 years ago
Convert 15*c in Fahrenheit
balandron [24]

Answer:

it is 59 degrees Fahrenheit

5 0
2 years ago
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Because light travels in a straight line and casts a shadow, Isaac Newton hypothesized that light is
KonstantinChe [14]

A stream of particles

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3 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
3 years ago
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