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erastovalidia [21]
3 years ago
12

Types of mechanical waves include

Physics
1 answer:
USPshnik [31]3 years ago
4 0
I think surface waves
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What do you mean? I'm confused... You need to put the rest  of the question

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3 years ago
John throws a baseball from a standing position. Which of the three possible paths will the ball follow?
Marysya12 [62]

Answer: C

Explanation:

Bro this is common sense. Throw something across the room and watch it. It isnt going to be repulsed and fly up. It also isnt going to continue on forever. It will clearly fall and hit the ground

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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
Dimensionally, which of the following could be a velocity? *
astra-53 [7]

Answer:

5 meters per second

Explanation:

5m is the distance

5m west is the vector

5m per second is the velocity

5m per second west is unknown

5 0
3 years ago
Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle
marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

Snell's law establishes that reflection is subject to

n_1sin\theta_1 = n_2sin\theta_2

Where,

\theta = Angle between the normal surface at the point of contact

n = Indices of refraction for corresponding media

The total internal reflection would then be given by

n_1 sin\theta_1 = n_2sin\theta_2

(1.54) sin\theta_1 = (1.33)sin(90)

sin\theta_1 = \frac{1.33}{1.54}

\theta = sin^{-1}(\frac{1.33}{1.54})

\theta = 59.72\°

Therefore the \alpha_{max} would be equal to

\alpha = 90\°-\theta

\alpha = 90-59.72

\alpha = 30.27\°

Therefore the largest value of the angle α is 30.27°

3 0
3 years ago
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