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Nezavi [6.7K]
3 years ago
10

What can you conclude about the total mechanical energy of a pendulum as it swings back and forth?

Physics
1 answer:
Alchen [17]3 years ago
8 0

Answer:

The total mechanical energy of a pendulum is conserved neglecting the friction.

Explanation:

  • When a simple pendulum swings back and forth, it has some energy associated with its motion.
  • The total energy of a simple pendulum in harmonic motion at any instant of time is equal to the sum of the potential and kinetic energy.
  • The potential energy of the simple pendulum is given by P.E = mgh
  • The kinetic energy of the simple pendulum is given by, K.E = 1/2mv²
  • When the pendulum swings to one end, its velocity equals zero temporarily where the potential energy becomes maximum.
  • When the pendulum reaches the vertical line, its velocity and kinetic energy become maximum.
  • Hence, the total mechanical energy of a pendulum as it swings back and forth is conserved neglecting the resistance.
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A .140kg baseball traveling 35.0m/s strikes the catchers mit bringing the ball to rest, recoils backward 11.0cm what’s the avera
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Average force applied by the call on the glove = 780 N

Explanation:

mass of baseball=0.140 kg

initial velocity = Vi= 35 m/s

Final velocity=Vf= 0

distance traveled=11 cm= 0.11 m

using the kinematic equation Vf²= Vi²+ 2 a d

where a = acceleration

0²= 35²+ 2 a (0.11)

a=-5568.2 m/s²

Now force is given by F= ma

F= 0.140 (5568.2)

F=-780 N

The negative sign signifies that the force acts in the opposite direction.

so the average force= 780 N

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A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

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3 years ago
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