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3 years ago

Diffuse reflection occurs when the size of surface irregularities is

Physics

1 answer:

erik [133]3 years ago

7 0

Answer:

Varies from point to point on the surface of the material.

Explanation:

Diffusion reflection is the reflection of light due to irregular reflection angles on the material surface. It might be as a result of microscopic irregularities on the material surface or just plain non reflecting ability of the material.

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Comment on the statement "Sharper the curve, more is the bending."

Tatiana [17]

The direction of the angular velocity of the minute hand of a wall-clock is perpendicular to the wall and directed inwards.

The passenger slides away from the centre of the curve.

The angle θ which a cyclist should make with the vertical while taking a circular turn of radius r with velocity v given by

tan θ = v^2/rg

Sharper the curve, smaller is the radius and greater is the value of 0.

8 0

2 years ago

A conducting loop in the form of a circle is placed perpendicular to a magnetic field of 0. 50 t. If the area of the loop decrea

Kisachek [45]

The induced emf in the loop is -1500 μ V or - 0.0015 V .

According to the question

A conducting loop in the form of a circle is placed perpendicular to a magnetic field of 0. 50 t.

i.e

Magnetic field (B) = 0. 50 T

Area of circle or loop = $\pi r^{2}$

Now,

The area of the loop decreases at a rate of 3. 0 × 10⁻³ m/s

i.e

dA = 3. 0 × 10⁻³ meter²

dt = 1 sec

As per the formula of Induced e.m.f in the loop

emf is dependent on number of turns of coil, shape of the coil, strength of magnet and speed with which magnet is moved. Emf is independent of resistivity of wire of the coil.

$e=-B*\frac{dA}{dt}$

where A is the area of the loop.

Now ,

Substituting the values in the formula

$e=-B*\frac{dA}{dt}$

$e= - 0.50 *\frac{ 3 * 10^{-3}}{1}$

e = - 0.0015 V

e = -1500 * 10⁻⁶ V

e = -1500 μ V

Negative just signifies emf will such be induced that current induced will oppose change in magnetic field though it

To know more about induced emf here:

brainly.com/question/16764848

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5 0

2 years ago

Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v

vladimir1956 [14]

Answer:

$P(X\ge 1) = 0.9502$

Explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$\lambda = density$

$\lambda = \frac{3}{10}$

$\lambda = 0.3$

T = the light years

$T = 10$

Calculating $P(X \ge 1)$

In probability:

$P(X \ge 1) = 1 - P(X = 0)$

Calculating P(X=0)

Substitute 0 for k and the values for $\lambda$ and T in

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}$

$P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}$

$P(X=0) = (3)^0 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-3}$

$P(X=0) = e^{-3}$

$P(X=0) = 0.04979$

Substitute 0.04979 for P(X=0) in $P(X \ge 1) = 1 - P(X = 0)$

$P(X\ge 1) = 1 - 0.04979$

$P(X\ge 1) = 0.95021$

$P(X\ge 1) = 0.9502$ --- approximated

<em>Hence, the required probability is 0.9502</em>

3 0

3 years ago

A 41-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.5 m/s. Neglect air resistan

postnew [5]

The altitude of the pole vaulter as she crosses the bar is 5 m.

<h3>The altitude of the bar</h3>

v² = u² - 2gh

where;

v is final velocity of the pole vaulter
u is the initial velocity of the pole vaulter
h is altitude of the bar

h = (u² - v²)/2g

h = (10² - 1.5²)/(2 x 9.8)

h = 5 m

Thus, the altitude of the pole vaulter as she crosses the bar is 5 m.

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3 0

2 years ago

Suppose two objects are attracting each other gravitationally. If you double the distance between them, the strength of their gr

Roman55 [17]

Answer:

decreases by a factor of 4

Explanation:

Lets take mass of the objects are m₁ and m₂ and the distance between them at initial condition is R.

We know that attraction force between two object given as

$F=G\dfrac{m_1m_2}{R^2}$

G=Constant

If distance become 2 R

R' = 2R

$F'=G\dfrac{m_1m_2}{R'^2}$

$F'=G\dfrac{m_1m_2}{(2R)^2}$

$F'=G\dfrac{m_1m_2}{4R^2}$

$F'=\dfrac{1}{4}\times G\dfrac{m_1m_2}{R^2}$

$F'=\dfrac{F}{4}$

Therefore we can say that gravitational force will become one forth of the initial force.

6 0

3 years ago