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3 years ago

What is tan(16°)? A. 0.96 B.0.16 C.0.39 D.0.29

Physics

1 answer:

Alex Ar [27]3 years ago

5 0

Answer:

D. 0.29

Explanation:

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A student locates a double-slit assembly 1.40 m from a reflective screen. The slits are separated by 0.0572 mm. (a) Suppose the

astra-53 [7]

Complete Question

A student locates a double-slit assembly 1.40 m from a reflective screen. The slits are separated by 0.0572 mm.

(a) Suppose the student aims a beam of yellow light, with a wavelength of 589 nm, toward the slit assembly, and this makes an interference pattern on the screen. What distance (in cm) separates the zeroth-order and first-order bright fringes (a.k.a. maxima)?

(b)

Now suppose that blue light (with

λ = 415 nm)

is used instead. What distance (in cm) will now separate the second-order and fourth-order bright fringes?

Answer:

The distance of separation is $z_1 - z_o = 1.44cm$

The distance of separation is $z_4 - z_2 = 2.031cm$

Explanation:

From the question we are told that

The distance from the screen is $D = 1.40m$

The slit separation is $d = 0.0572 mm = 0.0572 *10^{-3} m$

The wavelength of the yellow light is $\lambda_y = 598nm$

The distance of a fringe from the central maxima is mathematically represented as

$z_n = n \frac{\lambda_y D}{d}$

Where n is the order of the fringe so the distance of separation between

The distance that separates first order from zeroth order bright fringe can be evaluated as

$z_1 - z_o = (1 - 0 ) \frac{\lambda_y D}{d}$

Substituting values

$z_1 - z_o = (1 - 0 ) \frac{590*10^{-9} 1.40}{0.0572 *10^{-3}}$

$z_1 - z_o = 0.0144m$

Converting to cm

$z_1 - z_o = 0.0144m = 0.0144*100 = 1.44cm$

The wavelength of blue light is $\lambda _b$

So the distance that separates second order from fourth order bright fringe can be evaluated as

$z_4 - z_2 = (4 - 2 ) \frac{\lambda_y D}{d}$

Substituting values

$z_4 - z_2 = (4 - 2 ) \frac{415*10^{-9} 1.40}{0.0572 *10^{-3}}$

$z_4 - z_2 = 0.02031 \ m$

Converting to cm

$z_4 - z_2 = 0.02031m = 0.02031*100 = 2.031cm$

7 0

3 years ago

A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the

Naya [18.7K]

The stone's acceleration, velocity, and position vectors at time $t$ are

$\mathbf a(t)=-g\,\mathbf j$

$\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j$

$\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j$

where

$g=9.80\dfrac{\rm m}{\mathrm s^2}$

$v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}$

$v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}$

and $y_i$ is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component $(\mathbf j)$ of the position vector to 5 m and solve for $y_i$ :

$5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2$

$\implies\boxed{y_i\approx70.8\,\mathrm m}$

(b) Evaluate the horizontal component $(\mathbf i)$ of the position vector when $t=6.1\,\mathrm s$ :

$\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}$

(c) The rock's velocity vector has a constant horizontal component, so that

$v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}$

where $v_{f,x}$

For the vertical component, recall the formula,

${v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y$

where $v_{i,y}$ and $v_{f,y}$ are the initial and final velocities, $a$ is the acceleration, and $\Delta y$ is the change in height.

When the rock hits the ground, it will have height $y_f=0$ . It's thrown from a height of $y_i$ , so $\Delta y=-y_i$ . The rock is effectively in freefall, so $a=-g$ . Solve for $v_{f,y}$ :

${v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)$

$\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}$

(where we took the negative square root because we know that $v_{f,y}$ points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

$\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which has a magnitude of

$\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}$

(d) The acceleration vector stays constant throughout, so

$\mathbf a(t)=\boxed{-g\,\mathbf j}$

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2 years ago

What is the force when two charged spheres distance is in half

slamgirl [31]

Answer:

When two spheres, each with charge Q, are positioned a distance Rapart, they are attracted to ... doubled, the electric-force between the two spheres

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3 years ago

Show that 1.0 m/s = 3.6 km/h . Hint: Show the explicit steps involved in converting 1.0 m/s = 3.6 km/h.

sergiy2304 [10]

Just multiply the "1.0 m/s" by ' 1 ' a few times. (Remember that a fraction with the same quantity on top and bottom is equal to ' 1 ' .)

(1.0 m/sec) · (1 km/1000 m) · (60 sec/min) · (60 min/hr) =

(1.0 · 60 · 60 / 1,000) (m · km · sec · min / sec · m · min · hr) =

(3,600 / 1,000) (km / hr) =

3.6 km/hr .

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3 years ago

If the distance between the two mass double what happens to the gravitational force

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If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

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3 years ago