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3 years ago

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A car is traveling north with a velocity of 18.1 m/s. Find the velocity of the car after 7.50 seconds if the acceleration is 2.4

m/s^2. *

1 answer:

Vedmedyk [2.9K]3 years ago

3 0

Hello!

Vx = V0x + Ax*t
Vx = 18.1 + 2.4t

Let’s take time as 7.50 seconds:
Vx = 18.1 + 2.4*7.50
Vx = 18.1 + 18 = 36.1 m/s

Then, the final velocity of the car is 36.1 m/s.

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You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 7.5 x

Blizzard [7]

Complete Question

You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced $7.5 * 10^{-3}$ meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?

Answer:

The wavelength is $\lambda = 56250 nm$

Explanation:

From the question we are told that

The distance of slit separation is $d = 7.5 *10^{-3} \ m$

The distance of the screen is $D = 4 \ m$

The distance between the bright spot closest to the center of the interference is $k = 1.5 \ cm = 0.015 \ m$

Generally the width of the central maximum fringe produced is mathematically represented as

$y = 2 * k = \frac{ D * \lambda}{d}$

=> $2 * 0.015 = \frac{ \lambda * 4}{ 7.5 *10^{-3}}$

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3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

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