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borishaifa [10]
3 years ago
11

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: -4000 kg • m/s

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3 years ago
A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car
dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

Normal force = Gravitational force - centripetal force

At the foot of a round hill

Normal Force = centripetal force + Gravitational force

4 0
3 years ago
During the 28-day lunar cycle, the positions of the Sun,
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The answer is full moon because that’s how it changes in life
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A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above
Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed v_{1}= 2\ m/s

Speed v_{2}=7\ m/s

Pressure in main pipe P_{1}=2\times10^{5}\ Pa

(I). We need to calculate the diameter

Using equation of continuity

Av_{1}=Av_{2}

\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}

(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7

d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}

d_{2}=5.345\ cm

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})

P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)

P_{2}=1.28500\times10^{5}\ Pa

(III).  If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}

P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})

Here, h_{3}=0

Put the value in the equation

P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17

P_{3}=3.5400\times10^{5}\ Pa

Hence, This is required solution.

7 0
3 years ago
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