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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:

You might be interested in

a spring with a constant of 80N/m is stretched by a force of 240N. how much the displacement of the spring from equilibrium?

Inessa05 [86]

Answer:

1200N/m

Explanation:

given parameters:

force on the motorcycle spring is 240N

Extension 2cm or 0.02m

unknown _

spring constant:

solution:

to a spring a force applied is given as :

f=ke

f is applied as force

k is spring constant

e is the Extension

240= kx0.02

k=1200N/m

8 0

3 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

3 years ago

WHOEVER GET IT RIGHT GETS 50 POINTS

AVprozaik [17]

Answer:

ANSWER BELOW I

Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in

m/s2

. For example, for Earth, 445 N = 45.4 × 9.8

m/s2

:Notice that the x-axis values will be gravity in

m/s2

, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).

3 0

4 years ago

What are the consequences of sediment pollution?

Nonamiya [84]

Loss of habitats for fish, birds, and other wildlife. Sediment pollution is one of the leading causes of the loss of the wetlands, but it’s not just the wetlands. Changes in the nutrients in your water. The same problem that affects the fish in your area may also affect you. Other drinking water contamination.

6 0

3 years ago

Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an

disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = $\frac{1}{2} *m*vf^{2}$ (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

$vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s$

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0

3 years ago