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4 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.

2 answers:

goblinko [34]4 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]4 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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The speed of an arrow fired from a compound

san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

$X=v_{ox}*t$

$v_{0x} =v_0cos(\alpha)$

Y axis:

$Y= Y_0 +v_{y0} t - \frac{g}{2} t^2$

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

Caculate the segcond degree equation to obtain the two possible values for t:

$t_1= 10.18 \\t_2= -0.04046$

But, in physics, time it could not be negative, so we take $t_1= 10.18$

Caculate now:

$X=79m/s*cos(\39)*10.18s= 624.996 m$

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

$v_0= 79m/s+13m/s= 92m/s$

Using the same procedure that item A, caculate X

First, we need to know the new time

$0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

And we obtain:

$t_1=11.845s\\t_2=-0.041s$

One more time, we take the positive time: $t_1=11.845s$

Finally:

$X=92m/s *cos(39)*11.845s=846.887 m$

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