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2 years ago

12

Un hamster esta sentado sobre un tocadisco cuya rapidez angular es constante si el hamster se mueve a un punto localizado al dob

le de la distancia del centro entonces su rapidez lineal...
a) sera la mitad
b) se duplica
c) se mantendra igual

1 answer:

kotegsom [21]2 years ago

8 0

Answer:

b) se duplica

Explanation:

The disk is moving with constant angular velocity, let's call it $\omega$ .

The linear velocity of a point on the disk is given by

$v=\omega r$

where r is the distance of the point from the axis of rotation.

In this problem, the object is moved at a distance twice as far as the initial point, so

$r' = 2r$

Therefore, the new linear velocity is

$v'=\omega r' = \omega (2r) = 2 \omega r = 2 v$

So, the velocity has doubled, and the correct answer is

b) se duplica

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Which is an example of current electricity? <br>

VARVARA [1.3K]

A. A child rubs a balloon

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2 years ago

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How

Vlad [161]

Answer:

769,048.28Joules

Explanation:

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How much energy was lost to air friction during this bump

The energy lost due to friction is expressed using the formula;

Energy lost = Potential Energy + Kinetic Energy

Energy lost = mgh + 1/2mv²

m is the mass

g is the acceleration due to gravity

h is the height

v is the speed

Substitute the given values into the formula;

Energy lost = 56(9.8)(1400) + 1/2(56)(5.10)²

Energy lost = 768,320 + 728.28

Energy lost = 769,048.28Joules

<em>Hence the amount of energy that was lost to air friction during this jump is 769,048.28Joules</em>

6 0

2 years ago

Beginning 145 miles directly south of the city of Hartville, a car travels due west. If the car is travelling at a speed of 42 m

ziro4ka [17]

Answer:

The rate of change of the distance is 14.89.

Explanation:

Given that,

Distance = 145 miles

Speed of car = 42 miles/hr

Distance covered by car = 55 miles

We need to calculate the the rate of change of the distance

According to figure,

Let OA is x, and AB is y.

Now, using Pythagorean theorem

$x^2=y^2+145^2$

On differentiating

$2x\dfrac{dx}{dt}=2y\dfrac{dy}{dt}$

$\dfrac{dx}{dt}=\dfrac{y}{x}\dfrac{dy}{dt}$

$\dfrac{dx}{dt}=\dfrac{55\times42}{\sqrt{55^2+145^2}}$

$\dfrac{dx}{dt}=14.89\ miles/hr$

Hence, The rate of change of the distance is 14.89.

8 0

3 years ago

The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.

Alexus [3.1K]

Answer:

(a) θ= 43.89°

(b) $v_{1} = 1.88\sqrt{3} \frac{m}{s}$

$v_{2} = 6.79 \frac{m}{s}$

Explanation:

Ball 1:

$u_{1} = 7.52\frac{m}{s}$

Ball 2:

$u_{2} = 0\frac{m}{s}$

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

$\frac{1}{2} m_{1}u_{1}^{2} = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}$

and

$m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

$u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$

and

$u_{1} = v_{1} + v_{2}$

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

$56.55 = v_{1} ^{2} + v_{2} ^{2}$

and

$u_{1} = v_{1}cos(30) + v_{2}cos(\theta)$

Solving we get:

$(\frac{2u_{1}-\sqrt{3}v_{1} }{2cos(\theta)})^{2} = 56.55-v_{1} ^{2}$

From conservation in y-direction, we get:

$0 = v_{1}sin(30) - v_{2}sin(\theta)$

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.

8 0

3 years ago

Please helppppppppppppp

artcher [175]

The ball's initial velocity must be 28.0 m/s

Explanation:

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path, consisting of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

First, we study the vertical motion of the ball: since it is a uniformly accelerated motion, we can use the suvat equation to find the time it takes for the ball to reach the ground,

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

where

s = 40 m is the vertical displacement, the height of the cliff (we chose downward as positive direction)

u = 0 is the initial vertical velocity of the ball

t is the time of flight of the ball

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

Solving for t, we find:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(40)}{9.8}}=2.86 s$

Now we can analyze the horizontal motion: since this is a uniform motion, the horizontal speed is constant, and it is given by

$v_x = \frac{d}{t}$

where:

d = 80 m is the horizontal distance covered by the ball

t = 2.86 s is the time of flight

Substituting,

$v_x = \frac{80}{2.86}=28.0 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago