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3 years ago

1. Is it possible for the ball to move so quickly that the angle between the cable and vertical post stays at ninety degrees?

Physics

1 answer:

sp2606 [1]3 years ago

5 0

Answer:

Tetherball is an interesting game in which two players tries to hit the ball hard so that it goes around the

pole.Each time the player hits the ball, it's orbit rises higher off the ground.Let's understand the physics

behind this.The motion of a tetherball is governed by two forces.These two forces combine to generate a

net force, i.e. centripetal force.If the ball is moving more quickly, it requires a greater centripetal force,

which in turn requires a greater tension force.Since the ball's weight hasn't changed, the angle of the

tension force changes until the ball is in vertical equilibrium.

To access this physics simulation visit: http://goo.gl/xVdwgO Page 02Exploration Series www.ck12.org

Ball Mass : This slider controls the mass of the ball. A ball with more mass will have more inertia, requiring

a greater net force to accelerate it. A ball with more mass will ALSO have a greater gravitational force

acting on it. Watch both of these effects occur when you manipulate this slider.

Cable Length : This slider controls the length of the cable. A longer cable is capable of allowing a greater

circular radius of motion for the ball. It is important to remember that the radius of the circular motion is

NOT equal to the length of the cable. Instead, if you want to understand the size of the circle of the ball's

motion, ignore the cable and just imagine the path of the ball.

Ball Speed : This slider controls the speed of the ball - imagine a kid just hit the ball and it sped up. A ball

moving more quickly is also accelerating more quickly because its velocity is changing as it moves in a

circle (remember that changes in DIRECTION of velocity 'count' as changes to velocity).

Force Diagram : This allows you to turn on or off the diagram of the forces acting on the ball. Look for the

ball to be in vertical force balance, which means the vertical component of tension is canceled by the

gravitational force. The ball should NOT be in horizontal force balance - it is accelerating towards the center

of the circle! It is important to note that this free body diagram should really be moving with the ball so that

To access this physics simulation visit: http://goo.gl/xVdwgO Page 03Exploration Series www.ck12.org

the tension force always points along the cord - we are just showing the forces at the moment the ball is at

the furthest-right on this screen.

Centripetal force vs Tetherball speed : This is a plot of centripetal force required to keep the ball in

circular orbit about the pole as a function of its speed. As expected, a more quickly-moving ball is changing

in velocity more often in a given amount of time, and so is accelerating more. This greater (centripetal, or

center-pointing) acceleration requires a greater net force.

Explanation:

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Answer:

B. An electric current into a magnetic field

Explanation:

The generation of electrical power requires relative motion between a magnetic field and a conductor. In a generator, mechanical energy is converted into electrical energy. The electricity produced by most generators is in the form of alternating current.

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3 years ago

Read 2 more answers

Um comentarista de futebol certa vez comentou:"A bola bateu na trave e voltou duas vezes mais forte". Sabendo que quando a bola

ryzh [129]

Answer:

Por ela ter batido na trave, não tem como voltar 2x mais forte, por que toda ação correspondente a uma reação de igual intensidade, mas que atua no sentido oposto

Explanation:

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2 years ago

Light travels 3,00,000 km/s . Is it velocity or speed?

Ludmilka [50]

Answer:

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Explanation:

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2 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

Learn more about accelerated motion:

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3 years ago

What is a gravitational field and how its strength be measured

Yakvenalex [24]

A gravitational field is the field generated by a massive body, that extends into the entire space. Every object with mass m experiences a force F when immersed in a gravitational field. The intensity of the force is equal to
$F= \frac{GM}{r^2} m$
where $G=6.67 \cdot 10^{-11} m^3 Kg^{-1} s^{-2}$ is the gravitational constant, M is the mass of the source of the field (e.g. the mass of a planet), and r is the distance between the object and the source of the field. The force is always attractive.

A possible way to measure the intensity of a gravitational field is by measuring the acceleration a of the object immersed in this field. In fact, for Newton's second law we have:
$F=ma$
but since
$F= \frac{GM}{r^2} m$
we can write
$a = \frac{GM}{r^2}$
Therefore, by measuring the acceleration of the object, we also measure the intensity of the field.

5 0

3 years ago