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tresset_1 [31]
2 years ago
13

A group of physics students hypothesize that for an experiment they are performing, the speed of an object sliding down an incli

ned plane will be given by the expression v=2gd(sin(θ)−μkcos(θ))−−−−−−−−−−−−−−−−−−√. For their experiment, d=0.725meter, θ=45.0∘, μk=0.120, and g=9.80meter/second2. Use your calculator to obtain the value that their hypothesis predicts for v.
Physics
2 answers:
goblinko [34]2 years ago
6 0

Answer:

v = 2.974

Explanation:

Perhaps the formula should be

v = √(2*g*d (sin(θ) - uk*cos(θ) )                    This is a bit easier to read.

v = √(2* 9.80*0.725(0.707 - 0.12*0.707) )   Substitute values. Find 2*g*d

v = √14.21 * (0.707 - 0.0849)                        Figure out Sin(θ) - uk cos(θ)  

v = √14.21 * (0.6222)

v = √8.8422                                                  Take the square root of the value

v = 2.974

Tju [1.3M]2 years ago
6 0

Answer:

v = 2.97 m/s

Explanation:

As we know that the velocity expression for the given experiment is

v = \sqrt{2gd(sin\theta - \mu_kcos\theta)}

now we know that

d = 0.725 m

\theta = 45 ^o

\mu_k = 0.120

g = 9.80

now we have

v = \sqrt{2(9.80)(0.725)(sin45 - 0.120cos45)}

v = 2.97 m/s

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Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
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The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
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Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

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