Classify the following situations as involving balanced or unbalanced forces.

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

sp2606 [1]

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:

5 0

3 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

6 0

3 years ago

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0

3 years ago

Find the momentum of a particl with a mass of one gram moving with half the speed of light.

joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

6 0

3 years ago

Why is the linear coefficient of thermal expansion important for a restorative material?

iren2701 [21]

Additionally, the coefficient of thermal expansion of the restorative material should be comparable to the coefficient of thermal expansion of the tooth structure, since a significant difference between the two could result in thermal-induced stress at the cavity wall and subsequent marginal failure.

This study's objective was to assess how thermal stress affected the marginal integrity of restorative materials with various adhesive and thermal characteristics. As an alternative to clinical trials, which are expensive and time-consuming, evaluation of restorative materials under laboratory simulations of clinical function is frequently carried out. Thermal cycling regimens, which are in vitro techniques that subject the restoration and the tooth to extremely high temperatures, are frequently used in laboratory simulations to replicate thermal stresses that naturally occur in vivo.

Learn more about Thermal Stress here-

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5 0

1 year ago