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Alinara [238K]
4 years ago
9

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at angles of 23.8 ∘ and 65.0 ∘, r

espectively, to that of the first. If unpolarized light is incident on the stack, the light has an intensity of 71.5 W/cm2 after it passes through the stack.
a) If the incident intensity is kept constant, what is theintensity of the light after it has passed through the stack if thesecond polarizer is removed?
Physics
1 answer:
GalinKa [24]4 years ago
4 0

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Cliff divers at Acapulco jump into the sea macias (fjm793) – Homework 3, 2d motion 19-20 – dowd – (WoffordWPHY11920 2) 4 from a
Stells [14]

Answer:

v = 7.67 m/s

Explanation:

Given data:

horizontal distance 11.98 m

Acceleration due to gravity 9.8 m/s^2

Assuming initial velocity is zero

we know that

h = \frac{gt^2}{2}

solving for t

we have

t = \sqrt{\frac{2h}{g}}

substituing all value for time t

t = \sqrt{\frac{2\times 11.98}{9.8}}

t = 1.56 s

we know that speed is given as

v = \frac{d}{t}

v =\frac{11.98}{1.56}

v = 7.67 m/s

7 0
3 years ago
If energy is absorbed because chemical reaction is ______.
melomori [17]
<h2>Answer:</h2>

The reaction is <u>D. endothermic</u>.

<h2>Explanation:</h2>

Absorption or need of energy from surroundings is called endothermic process. Heat is the major common form of energy which is absorbed in this process.  In this process, enthaply of reactant is less than the product.

For example, an ice cube placed in the hand for some time starts to melt and it takes energy in the form of heat from the hand. This process always requires energy because of the breaking of existing and formation of new chemical bonds.

4 0
3 years ago
1. Determina el momento que produce una fuerza de 7 N tangente a una rueda de un metro de diámetro, sabiendo que el punto de apl
Rudik [331]

Answer:

τ= F r     into the blade

Explanation:

The moment of a force is defined by

         τ = F x r

where the bold indicates vectors

Let us write in the expression in magnitude

         τ = F r sin θ

in our case the force is tangent to the wheel therefore the angle between F and the radius is 90º, and the sin 90 = 1

       τ= F r

The direction of τ can be used by the rule of the right hand, the fingers curve in the direction of the torque when advancing from the force to the radius and the thumb points in the direction of the torque.

In this case, for a clockwise rotation, the fingers are curved in the direction and the thumb points into the blade, this is the direction of the τ.

TRASLATE

El momento de una fura es definido por

         τ = F x r

donde la negrillas indican vectores

Escribamos en ta expresión en magnitud

          τ = F r sin θ

en nuestro caso la fuerza es tangente a la rueda por lo tanto el angulo entre F y el radios es 90º, y el sin 90=1

        τ = F r

la dirección de tau la podemos  usar la regla de la mano derecha, los dedos curva en la dirección del torque al avanzar dese la fuerza al radio y el pulgar apunta en la dirección del torque.

En este caso para un giro en sentido horario los dedos se curvan ente sentido y el pulgar apunta hacia dentro de lla hoja, esta es la dirección del troque

5 0
4 years ago
The cells of collenchyma have at the corners​
saw5 [17]

Explanation:

cellulose and pectin are present on the cells of collenchyma...

8 0
3 years ago
Two forces,
serg [7]

First compute the resultant force F:

\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N

\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N

\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N

Then use Newton's second law to determine the acceleration vector \mathbf a for the particle:

\mathbf F=m\mathbf a

(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a

\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}

Let \mathbf x(t) and \mathbf v(t) denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so \mathbf v(0)=0. Then the particle's velocity vector at <em>t</em> = 10.4 s is

\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du

\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}

\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}

If you don't know calculus, then just use the formula,

v_f=v_i+at

So, for instance, the velocity vector at <em>t</em> = 10.4 s has <em>x</em>-component

v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}

(b) Compute the angle \theta for \mathbf v(10.4\,\mathrm s):

\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ

so that the particle is moving at an angle of about 313º counterclockwise from the positive <em>x</em> axis.

(c) We can find the velocity at any time <em>t</em> by generalizing the integral in part (a):

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

Then using the fundamental theorem of calculus again, we have

\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du

where \mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m is the particle's initial position. So we get

\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du

\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}

\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m

So over the first 10.4 s, the particle is displaced by the vector

\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

3 0
3 years ago
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