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Gemiola [76]
3 years ago
10

Please help. I will give brainliest to first good answer

Physics
2 answers:
Arturiano [62]3 years ago
7 0
Answer: C). Light rays are bent by the water in the glass

It refracts the light
The bending of a light moving from one end to another is called refraction
Elza [17]3 years ago
6 0
The answer is c i had this lesson before :)
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Neutrally charged!!!!!!!!!!!!!!!!!!!!!

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A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.
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Answer:

W = 95.8J

Explanation:

Given

S = 6.00m

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A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-
Snezhnost [94]

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Explanation:

E=kq/r

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Which describes sublimation?
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2 years ago
What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

5 0
3 years ago
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