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4 years ago

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Recall some things you already know about projectile motion. Does a force in the vertical direction affect the horizontal compon

ent of an object’s velocity? In this situation, can you use the horizontal velocity component to find the time required to travel some horizontal distance?

1 answer:

azamat4 years ago

7 0

Answer:

vertical force cannot change the velocity on the x-axis. t =x/v₀ₓ

Explanation:

The force is a vector magnitude, so the forces on the x-axis affect the acceleration on this axis. Consequently a vertical force cannot change the velocity on the x-axis.

$F_{y}$ = m g

Fₓ = 0

The horizontal velocity in projectile motion is constant, if we neglect the air resistance, so it can be used to find the time of a horizontal displacement

x = v₀ₓ t

t =x/v₀ₓ

The only magnitude that is the same for both movements is the time that is a scalar

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A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i

Tresset [83]

Answer:

$\theta_1 = 0.400^o$

$\theta_2 =0.378^o$

Explanation:

From the question we are told that

The number of slits per cm is k = $161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m$

The order of the maxima is n = 1

The wavelength are $\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m$

The spacing between the slit is mathematically represented as

$d = \frac{ 0.01}{k}$

=> $d = \frac{ 0.01}{161}$

=> $d = 6.211 *10^{-5} \ m$

Generally the condition for constructive interference is

$n\lambda = d \ sin \theta$

At $\lambda_1$

$\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_1 = 0.400^o$

At $\lambda_2$

$\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_2 =0.378^o$

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3 years ago

An illustration of the major tectonic plates is provided below.

Julli [10]

Answer:

Plate A

Explanation: It has the continent of North America inside of it.

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3 years ago

Read 2 more answers

4. A latch holding a cart in place, hanging high above the ground, breaks and the cart falls. The cart started from rest and is

scoray [572]

Answer:

a) 31.4 m/s

b) 50.2 m

Explanation:

a) When an object is free falling, its speed is determined by the gravity force giving it acceleration. Equation for the velocity of free fall started from the rest is:

v = g • t

g - is gravitational acceleration which is 9.81 m/s^2, sometimes rounded to 10

t - is the time of free fall

So:

v = 9.81 m/s^2 • 3.2

v = 31.4 m/s ( if g is rounded to 10, then the velocity is 10 • 3.2 = 32 m/s)

b) To determine the distance crossed in free fall we use the equation:

s = v0 + gt^2/2

v0 - is the starting velocity (since object started fall from rest, its v0 is 0)

s = gt^2/2

s = 9.81 m/s^2 • 3.2^2 / 2

s = 50.2 m (if we round g to 10 then the distance is 10 • 3.2^2/2 = 51.2 meters)

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3 years ago

The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo

a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

$t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s$

The tire angular velocity before stopping is:

$\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s$

Also its angular decceleration:

$\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2$

Using the following equation motion we can findout the angle it makes during the deceleration:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega$ = 0 m/s is the final angular velocity of the car when it stops, $\omega_0$ = 114rad/s is the initial angular velocity of the car $\alpha$ = 14.75 rad/s2 is the deceleration of the can, and $\Delta \theta$ is the angular distance traveled, which we care looking for:

$-114^2 = 2*(-14.75)*\Delta \theta$

$\Delta \theta = 440rad$ or 440/2π = 70 revelutions

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3 years ago