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3 years ago

Describe how refraction concentrates wave energy so that some parts of a beach erode more.

1 answer:

3 0

Refraction is the bending of the waves which is result of the fact that different parts of the wave reach the water with different speeds because of the angle approaching the shore.

<span>The wave refraction disperses the wave energy in quiet water areas and sand is deposited.<span> </span></span>

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A car has a mass of 1000 kg and accelerates at 2 m/s2. What net force is exerted on the car?

prisoha [69]

Answer:

1000N

Explanation:

Based on force=mass*acceleration, if the acceleration is constant at 2 metres per second squared, 1,000kg*2m/s^2=2,000N of force.

If the acceleration steadily increases to 2m/s^2 in 20 seconds, take the average which is 1m/s^2 therefore force=1,000N

6 0

3 years ago

Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 11 cm that is placed in a spatially uniform m

Naddik [55]

Answer:

Explanation:

Given that,

Number of turn N = 40

Diameter of the coil d= 11cm = 0.11m

Then, radius = d/2 = 0.11/2 =0.055m

r = 0.055m

Then, the area is given as

A =πr²

A = π × 0.055²

A = 9.503 × 10^-3 m²

Magnetic Field B = 0.35T

Magnetic field reduce to zero in 0.1s, t = 0.1s

so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).

E.M.F is given as

ε = —N • dΦ/dt

Where magnetic flux is given as

Φ = BA

Then, ε = —N • dΦ/dt

ε = —N • dBA/dt

ε = —NBA/t

Then, its magnitude is

ε = NBA/t

Inserting the values of N, B, A and t

ε = 40×0.35×9.503×10^-3/0.1

ε = 1.33 V

Then, using the relationship between Electric field and electric potential

V = Ed

ε = E•d

E = ε/d

E = 1.33/0.11

E = 12.09 V/m

7 0

4 years ago

The accompanying table shows measurements of the Hall voltage and corresponding magnetic field for a probe used to measure magne

aalyn [17]

0.125 mm . is the thickness of the sample.

<h3>What do you mean by hall voltage ?</h3>

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

lof4

First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

Δ $V_{H} =\frac{I}{nqt} B$

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ $\frac{V}{T}$

b = —0.02 μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH = $100X10^{-6}$ V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

$a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26} } \\t=0.125mm$

To learn more about the hall voltage , Visit: brainly.com/question/19130911

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8 0

2 years ago

What are some alike things between the ozone layer and the greenhouse effect

umka2103 [35]

The ozone layer traps heat from the sun's heat. only three-fourths are reflected back out into space by the ozone layer. the greenhouse effect traps carbon dioxide and so does the ozone layer.

6 0

3 years ago

Peak wavelength of light coming from a star with a temp of 4300k

PSYCHO15rus [73]

from Wein's law

$\frac{k}{ \lambda_{max} } = T \\ \lambda_{max} = \frac{k}{T} \\ \lambda_{max} = \frac{3.898 \times {10}^{ - 3} }{4300} \\ = 9.065 \times {10}^{ - 7}$

3 0

3 years ago