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3 years ago

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An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c

onstant acceleration, it travels a distance x. How far would the automobile travel in coming to rest if it had the same acceleration but an initial speed 2v ?

1 answer:

Molodets [167]3 years ago

3 0

Answer:

4x

Explanation:

Use $v^{2} = u^{2} +2as$ to do the question.

For first instance,

$0 = v^{2} +2ax$ -------------------( 1 )

for second instance,

$0 = (2v)^{2} +2as$ -----------------( 2 )

So by (1) and (2),

s = 4x

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Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

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3 years ago

Describe two experiments to determine the speed of propagation of a transverse wave on a rope. You have the following tools to u

AnnZ [28]

Answer:

#See solution for details.

Explanation:

1.

Tools: $stopwatch, \ meter \ stick, \ mass \ measuring \ scale , \ force \ measuring \ device$ .

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-Attach one end of rope to a wall or post, shake from the unfixed end to generate a pulse. Measure the the time it takes for the pulse to reach the wall once it starts traveling using the stopwatch.

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$v=\frac{L}{t}$

$Experiment \ 2$ : Apply force of known value to the rope then use the following relation equation to find the speed of a pulse that travels on the rope.

$v=\sqrt{\frac{F}{\mu}}\ ,\mu=\frac{m}{L}$

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a runner runs 2.88 m/s north. she accelerates .350 m/s ^2 at a -52.0 degree angle at the point in the motion where she is runnin

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Answer:

uh

Explanation:

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3 years ago

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

A uniform motion along the horizontal (x) direction
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We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

$u_y$ is the initial vertical velocity of the ball, which is given by

$u_y = u sin \theta$

where

u = 23.5 m/s is the initial velocity

$\theta=33.5^{\circ}$ is the angle of projection

Substituting,

$u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

Substituting everything into the equation we get:

$-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0$

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

$v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

And substituting t = 3.59 s, we find

$d=(19.6)(3.59)=70.4 m$

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

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The general kinematic equations of motion for vertical displacement can also be simplified significantly. Write the simplified

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Answer:

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Explanation:

The general kinematic equation for horizontal displacement is gives as:

Δx = v₀t + (1/2)at²

Where

Δx = change in the x direction

v₀ = initial velocity

t = time

a = acceleration

If the body is vertically instead of horizontally, Δx is changed to Δy

Δy = v₀t + (1/2)at²

For a vertical moving body, the acceleration it experiences is the gravitational accerelation of the earth 'g'

So the equation becomes:

Δy = v₀t + (1/2)gt²

where g = 9.81 m/s if the body is moving downwards and g = -9.81 m/s if the body is moving upwards

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3 years ago