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dimulka [17.4K]
3 years ago
11

Hola tengo un taller de física y nose como resolver este pregunta

Physics
1 answer:
Blizzard [7]3 years ago
8 0

a) 0.26 h

b) 71.4 km

Explanation:

a)

In order to solve the problem, we have to know what is the final velocity of the car.

Here, we assume that the final velocity reached by the car is

v=300 km/h

Therefore, we can find the time taken by the car to reach this velocity by using the suvat equation:

v=u+at

where:

u = 250 km/h is the initial velocity

a=190 km/h^2 is the acceleration of the car

v = 300 km/h is the final velocity

t is the time

Solving for t, we find:

t=\frac{v-u}{a}=\frac{300-250}{190}=0.26 h

b)

In order to find the distance covered by the car, we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where:

s is the distance covered

u is the initial velocity

a is the acceleration

t is the time

For the car in this problem, we have:

u = 250 km/h

t = 0.26 h (calculated in part a)

a=190 km/h^2

Therefore, the distance covered is

s=(250)(0.26)+\frac{1}{2}(190)(0.26)^2=71.4 km

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Why is it a good idea to start with room temperature water in the calorimeter?
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Explanation:

It is a good idea to start with room temperature water in the calorimeter because the room temperature water helps to determine the heating up/cooling down because of the environment as the experiment takes place. Because the calorimeter heat is the same as the heat of the water.

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In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a
matrenka [14]

Answer:

Approximately \displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right].

Explanation:

Consider this 45^{\circ} slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin (0, 0).

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at 45^{\circ} to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as y = -x.

Convert the initial speed of this diver to SI units:

\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}.

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at g (g \approx \rm -9.81\; m\cdot s^{-2} near the surface of the earth.) At t seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

  • x-coordinate: 30.556t meters (constant velocity;)
  • y-coordinate: \displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2} meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate t from this expression, solve the equation between t and x for t. That is: express t as a function of x.

x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}.

Replace the t in the equation of y with this expression:

\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}.

Plot the two functions:

  • y = -x,
  • \displaystyle y= -0.0052535\;x^{2},

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of y (right-hand side of the equation, the part where y is expressed as a function of x.)

-0.0052535\;x^{2} = -x,

\implies x = 190.35.

The value of y can be found by evaluating either equation at this particular x-value: x = 190.35.

y = -190.35.

The position vector of a point (x, y) on a cartesian plane is \displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]. The coordinates of this skier is approximately (190.35, -190.35). The position vector of this skier will be \displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]. Keep in mind that both numbers in this vectors are in meters.

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3 years ago
A flat uniform circular disk (r= 2.00m,
dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum  

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

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ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is  

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

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3 years ago
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padilas [110]

Answer:

The Hubble space telescope.

Explanation:

Hubble is a telescope that observers the sky 24/7 non-stop, which means that for every day of the year it would have made a significant discovery, which of course includes your birthday. Furthermore, you can actually go to NASA website and find out what discovery was made on your birthday! This shows both the vastness of the universe <em>(it really has to be huge for a telescope to have a discovery for each day of the year!) </em> and the ceaseless work of the telescope!

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