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dedylja [7]
3 years ago
9

At what point in its motion is the kinetic energy of the end of a pendulum greatest? At what point is its potential energy great

est when it's kinetic energy is half its greatest value how much potential energy did it gain
Physics
2 answers:
Step2247 [10]3 years ago
8 0

-- The kinetic energy of any object is greatest when it's moving the fastest.
For the weight at the end of a pendulum, that's the instant when it passes
through the bottom of the arc.  That's the same place it would be if it were
not swinging at all.

-- An object's potential energy is greatest when it's in the highest position
of all.  For the weight at the end of a pendulum, that's the instant when it's
at either end of the arc, momentarily standing still and ready to reverse its
direction.  That's the height to which you lifted it, just before you let it go
to start swinging.

-- The potential and kinetic energy of an ideal pendulum keep trading off,
and their sum is always the same total.  It's exactly the amount of energy
you put into the pendulum when you lifted the weight up to some height
and let it go. 

As it falls from that height, it loses some potential energy because it's
not as high, and it gains some kinetic energy because it's moving faster. 
The amount it gains and the amount it loses are equal, and the sum of
the potential and kinetic energy never changes. 

Whenever the kinetic energy has reached half of its greatest value, the
potential energy has lost half of its greatest value.  That means that this
must happen when the weight is at half of its greatest height, midway (in
height) between the bottom and the ends of the arc.


DanielleElmas [232]3 years ago
7 0
The kinetic energy reaches a max once the mass returns to its equilibrium position. At this point all of the potential energy which is greatest at its maximum displacement from the center, turns into kinetic energy. 
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A piston motion moves a 25-kg hammerhead vertically down 1 m from rest to a velocity of 50 m/s in a stamping machine. What is th
Alja [10]

Answer:

31.005 KJ

Explanation:

We are given that

Mass of hammerhead=25 kg

Initial velocity,u=0

Final velocity,v=50 m/s

h=-1 m

h'=0

We have to find the change in total energy of the hammerhead.

Change in total energy=E_2-E_1=m(u_2-u_1)+\frac{1}{2}m(v^2-u^2)+mg(h-h')

Using the formula

Change in internal energy of hammerhead=m(u_2-u_1)=0

Change in total energy=\frac{1}{2}(25)(50)^2+25\times 9.8(-1)

Where g=9.8m/s^2

Change in total energy=31005 J=\frac{31005}{1000}=31.005 KJ

1 KJ=1000 J

3 0
3 years ago
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Fynjy0 [20]

Answer:

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Explanation:

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3 0
3 years ago
How many poles do you expect to see in your magnet? Look around your room. Make a list of 5 items that might be attracted to the
jeyben [28]

Answer:

1.1 Two poles: North and South Poles.

1.2 - Staple pin - Nail - Tip of my phone charger - Metal keys - Cloth Hanger

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5 0
3 years ago
The carnival ride has a 2.0m radius and rotates 1.1 times persecond.
Lady_Fox [76]

Answer:

a) v = 13.8 m / s , b) a = 95.49 m / s² , c) a force that goes to the center of the carnival ride  and d)   μ = 0.10

Explanation:

For this exercise we will use the angular kinematics relationships and the equation that relate this to the linear kinematics

a) reduce the magnitudes to the SI system

     w = 1.1 rev / s (2pi rad / 1rev) = 6.91 rad / s

The equation that relates linear and angular velocity is

     v = w r

     v = 6.91  2

     v = 13.8 m / s

b) centripetal acceleration is given by

     a = v² / r = w² r

     a = 6.91² 2

     a = 95.49 m / s²

c) this acceleration is produced by a force that goes to the center of the carnival ride

d) Here we use Newton's second law

     fr -W = 0

     fr = W

     μ N = mg

Radial shaft

      N = m a

      N = m w² r

      μ m w²  r = m g

      μ = g / w² r

      μ = 9.8 / 6.91² 2

      μ = 0.10

3 0
4 years ago
How fast must a 2.70-g ping-pong ball move in order to have the same kinetic energy as a 145-g baseball moving at 31.0 m/s
Helen [10]

Answer:

227 m/s

Explanation:

Kinetic energy formula:

  • \displaystyle \text{KE} = \frac{1}{2} mv^2
  • where m = mass of the object (kg)
  • and v = speed of the object (m/s)

Let's find the kinetic energy of the 145-g baseball moving at 31.0 m/s.

First convert the mass to kilograms:

  • 145-g → 0.145 kg

Plug known values into the KE formula.

  • \displaystyle \text{KE} = \frac{1}{2} (.145)(31.0)^2
  • \displaystyle \text{KE} = 69.6725 \ \text{J}

Now we want to find how fast a 2.70-g ping pong ball must move in order to achieve a kinetic energy of 69.6725 J.

First convert the mass to kilograms:

  • 2.70-g → 0.00270 kg

Plug known values into the KE formula.

  • \displaystyle 69.6725 = \frac{1}{2} (.00270)v^2
  • \displaystyle \frac{2(69.6725)}{.00270} =v^2
  • 57609.25926=v^2
  • v=227.1767137

The ping-pong ball must move at a speed of 227 m/s to achieve the same kinetic energy as the baseball.

4 0
3 years ago
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