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Delicious77 [7]
3 years ago
13

A dielectric material such as paper is inserted between the plates of a capacitor as the capacitor holds a fixed charge on its p

lates. What happens to the electric field between the plates as the dielectric is inserted?
Physics
1 answer:
Vikki [24]3 years ago
7 0

Answer:

Introducing a dielectric into a capacitor decreases the electric field, which decreases the voltage, which increases the capacitance.

Explanation:

A dielectric (or dielectric material) is an electrical insulator that can be polarized by an applied electric field. When a dielectric material is placed in an electric field, electric charges do not flow through the material as they do in an electrical conductor but only slightly shift from their average equilibrium positions causing dielectric polarization

Types of dielectric material

Ceramic, Mica paper glass

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Describe the differences among ultraviolet waves, visible light waves, and infrared waves. how are these waves alike?
sergeinik [125]
Our eyes are detectors which are designed to detect visible light waves (or visible radiation). ... The electromagnetic spectrum includes gamma rays, X-rays, ultraviolet, visible, infrared, microwaves, and radio waves. The only difference between these different types of radiation is their wavelength or frequency.
4 0
3 years ago
In order to create a charged object, you need to ________.
Brums [2.3K]
In order to create a charged object you need to transfer electrons either away or to the object by induction, conduction, or friction

Induction is without contact(like bringing a charged object to a electroscope charges the leaves at the bottom)
Conduction is with contact(like the previous answer, wore touching transfers the charge from a source to the object)
Friction(rubbing a balloon on wool)

7 0
2 years ago
Please answer the number 5 6 and 7. I don't know what to do its our hw in physics. (new lesson as well)
Aleks04 [339]

From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.

Just as you did with #4, please sketch these on paper
as I walk you through the solutions.  That'll help you see
immediately what's going on.

#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers  (3 m/s) x (4 sec) = 12 meters.

Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.

The total distance he covers is  (12m + 10m) = 22 meters.

#5.c).
Average speed (scalar)

                           = (distance covered)/(time to cover the distance)

                           = (22 meters)/(6 sec) = 3-2/3 m/s .

#5.d).
Displacement (vector)

                       = distance between the start-point and the end-point,
                          regardless of the route traveled,
                      
  in the direction from the start-point to the end-point.

Distance from the start-point to the end-point =

               √(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters

in the direction of  arctan(10/12) south of east

                             =  39.8° south of east.
 
#5.e).
Average velocity (vector) =

             (displacement vector) / (time)

         =  15.62 meters directed 39.8° south of east / 6 seconds

         =  2.603 m/s directed 39.8° south of east.

 #6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.

Direction = arctan(5.2/2.1) south of east

                =   68° south of east  =  158° bearing . 

#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)

                                                                         =  69.07 m/s .

Direction = arctan (57/39) south of west

               =   55.6° south of west

                    Bearing = 214.4°

                    Compass: 0.65° past "southwest by south".  


I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.

8 0
3 years ago
Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Ho
lidiya [134]

Answer:\frac{r_1}{r_2}=1.565

Explanation:

Given

two holes are made with different sizes

Hole 1 is large in size and hole 2 is small

If the volume flow rate of water is same for both the hole then small hole must be below the large hole because for same flow rate, velocity of water is large while cross-sectional area is small so it compensate to give same flow for both the holes.

Now for radius apply Bernoulli's theorem at hole 1 and 2

P_1+\rho gh_1=P_{atm}+\frac{1}{2}\rho v_1^2

P_2+\rho g6h_2=P_{atm}+\frac{1}{2}\rho v_2^2

if hole 1 is h distance below water surface then h_2=6h

and P_1=P_2=P_{atm}

Also v_1=\sqrt{2gh}

v_2=\sqrt{2g(6h)}

and Q=A_1v_1=A_2v_2

A=\pi r^2

thus \dfrac{r_1}{r_2}=\sqrt{\dfrac{v_2}{v_1}}

\dfrac{r_1}{r_2}=\sqrt{\dfrac{\sqrt{6h}}{\sqrt{h}}}

\frac{r_1}{r_2}=1.565

5 0
3 years ago
Cal ulate a moment of force o. 50 meter distance and 10 newton force​
lina2011 [118]

Answer:

500N/M

Explanation:

given that

force=10N

distance=50M

moment=force*distance

=10×50=500j

8 0
3 years ago
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