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Serjik [45]
3 years ago
5

A runner circles the track exactly 2 times for distance of 800m. it takes 4.0min. what is her average speed in m/s? what is her

average velocity?
Physics
1 answer:
Sindrei [870]3 years ago
5 0

If the runner ends where she started, then her total displacement is \Delta x=0\,\mathrm m, which means her average velocity will be

\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{0\,\mathrm m}{4.0\,\mathrm{min}\cdot\frac{60\,\mathrm s}{1\,\mathrm{min}}=0\,\dfrac{\mathrm m}{\mathrm s}

The total distance she traversed, however, was \Delta d=800\,\mathrm m, which gives her an average speed of

\bar s=\dfrac{\Delta d}{\Delta t}=\dfrac{800\,\mathrm m}{4.0\,\mathrm{min}\cdot\frac{60\,\mathrm s}{1\,\mathrm{min}}=3.3\,\dfrac{\mathrm m}{\mathrm s}

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Given a block of glass that is semi-circular, a laser pointer, protractor/ruler, and index card, design your own experiment to o
Charra [1.4K]

Answer:

we go up the ramp there is a point where the beam is reflected inside the block, we carefully step back to the point where the beam is horizontal, we measure this angle which is our critical angle.

Explanation:

To design the experiment of measuring the critical angle, we describe the phenomenon, when the light passes from a medium with a higher refractive index to one with a lower index, it separates from the normal one and the Critical Angle is defined as the Angle for which the refraction occurs at 90º

            n₂ sin θ₂ = n₁ sin 90

           n₁ / n₂ = sin θ₂

As we can see, we have to measure the angle with which the laser touches the exit surface of the glass block.

Design of the experiment:

    We place the glass block on the ramp and at the top we hit the conveyor for half the angle, we climb the block on the ramp and see that the angle of incidence of lightning on the exit face changes, part of the beam comes out of the glass , we see it by dispersion in the particles of dirty in the air; Maybe the conveyor or the laser should be moved slightly so that the beam touches the point of origin on the conveyor.

   

   When we go up the ramp there is a point where the beam is reflected inside the block, we carefully step back to the point where the beam is horizontal, we measure this angle which is our critical angle.

5 0
3 years ago
Read 2 more answers
What in th max age of a dog
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Ok, SO my family runs a Dog Day Care. ( littlecanineclubhouse.com if you need proof.)

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Such as a PUG, they can live anywhere between 15 to 20 years.
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6 0
3 years ago
What two factors affect the rate of acceleration of an object?
Masja [62]

For help with this answer, we look to Newton's second law of motion:

     Force = (mass) x (acceleration)

Since the question seems to focus on acceleration, let's get
'acceleration' all alone on one side of the equation, so we can
really see what's going on.

Here's the equation again:

                                                 Force = (mass) x (acceleration)

Divide each side by 'mass',
and we have:                            Acceleration = (force) / (mass) .

Now the answer jumps out at us:  The rate of acceleration of an object
is determined by the object's mass and by the strength of the net force
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5 0
3 years ago
A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
FrozenT [24]

Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

v=\omega r(2)

Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
3 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
2 years ago
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