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What is the amplitude and wavelength of the wave shown below?

masha68 [24]

Answer:

Option B

Explanation:

I did a question like this a long time ago so i'm not sure if its right so sorry if its wrong

3 0

2 years ago

4 points

zubka84 [21]

(a) 25lx

(b) 11.11lx

Explanation:

Illuminance is inversely proportional to the square of the distance.

So,

$I = k\frac{1}{r^2}$

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

$I_2 = k\frac{1}{(2r)^2}$

Dividing both the equation we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4} = 25lx$

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

$I_2 = k\frac{1}{(3r)^2}$

Dividing equation 1 and 3 we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9} = 11.11lx$

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0

3 years ago

Consider that 168.0 J of work is done on a system and 305.6 J of heat is extracted from the system. In the sense of the first la

Ilia_Sergeevich [38]

To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

$dQ = dU + dW$

Where,

Q = Heat

U = Internal Energy

By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore

$W = -168J \righarrow$ Work is done ON the system

$Q = -305.6J \rightarrow$ Heat is extracted FROM the system

Therefore the value of the Work done on the system is -158.0J

3 0

3 years ago

You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring th

AfilCa [17]

Answer:

$\partial \theta = 0.003$

Explanation:

we know that

$sin\theta = \frac{3.8}{146.4}$

$\theta = sin^{-1} \frac{3.8}{146.4}$

$\theta = 1.484°$

$\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians$

as we see that $sin\theta = \theta$

relative error $\frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}$

Where X_1 IS HEIGHT OF ROCK

$X_2$ IS THE HEIGHT OF ROAD

$\partial X$ = uncertainity in measuring distance

$\partial X = 0.05$

Putting all value to get uncertainity in angle

$\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}$

solving for $\partial \theta$ we get

$\partial \theta = 0.003$

3 0

3 years ago

In an inelastic collision a 2.5 kg ball moving at 7.5 m/s is caught by a 70kg man while the man is standing on ice. What is the

MrRa [10]

The velocity of the ball and the man is 0.259 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$