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avanturin [10]
3 years ago
14

A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in t

he pitcher's hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?
Group of answer choices
A) 10.5 rad/s
B) 0.0952 rad/s
C) 0.00857 rad/s
D) 117 rad/s
Physics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

D) 117 rad/s

Explanation:

We can treat this system as a circular motion where the origin is the elbow joint, the ball rotation velocity v is 35 m/s, the rotation radius is r = 0.3m.

As the ball is leaving the pitcher hand at such speed and rotation radius. Its angular velocity is:

\omega = \frac{v}{r} = \frac{35}{0.3} = 117 rad/s

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kondor19780726 [428]

The experiments that claimed to demonstrate cold fusion were found
to have been faulty by others who reviewed them.  Also, nobody else
was able to reproduce the finding in other laboratories.  In the world
of Science, this pretty much says that the initial claims were unfounded.

5 0
3 years ago
Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2
wolverine [178]

The particles can undergo small oscillations around x₂.

The given parameters;

  • <em>initial energy of the particles = E₁</em>
  • <em>final energy of the particles, E₂ = 0.33E₁</em>

The movement of the particles depends on the kinetic energy of the particles.

When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.

However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.

  • The maximum position the particle can oscillate is x₅
  • The half position the particles can oscillate is x₃

Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.

Thus, we can conclude that the particles can undergo small oscillations around x₂.

Learn more here:brainly.com/question/23910777

3 0
2 years ago
When a magnetic field is first turned on, the flux through a 20-turn loop varies with time according to Φm=5.0t2−2.0t, where Φm
goldfiish [28.3K]

Answer:

A) ( - 200t + 40 ) volts

B)  b) anticlockwise ,  c) anticlockwise , d) clockwise ,  e) clockwise

Explanation:

Given data:

magnetic flux (Φm) = 5.0t^2 − 2.0t

number of turns = 20

<u>a) determine induced emf </u>

E = - N \frac{d\beta }{dt}

  =  - N ( 10t - 2 ) = - 20 ( 10t - 2 )

  =  - 200t + 40  volts

<u>b) Determine direction of induced current </u>

i) at t = 0

 E = - 0 + 40  ( anticlockwise direction )

ii) at t = 0.10

E = -20 + 40 =  20 ( anticlockwise direction )

iii) at t = 1

E = - 200 + 40 = - 160 ( clockwise direction)

iv) at t = 2

E = -400 + 40 =  - 360 ( clockwise direction )

8 0
3 years ago
plz i need help fast................a motion along a straight line with a constant increase in velocity is ?​
Monica [59]

Answer:

uniform acceleration

Explanation:

The definition for uniform acceleration is:

if an object travels in  a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration is said to be uniform.

Hope this helps.

Good Luck

8 0
3 years ago
Read 2 more answers
A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input
KATRIN_1 [288]

Answer:

The peak-to-peak ripple voltage = 2V

Explanation:

120V and 60 Hz is the input of an unfiltered full-wave rectifier

Peak value of  output voltage = 15V

load connected = 1.0kV

dc output voltage = 14V

dc value of the output voltage of capacitor-input filter

where

V(dc value of output voltage) represent V₀

V(peak value of output voltage) represent V₁

V₀ = 1 - ( \frac{1}{2fRC})V₁

make C the subject of formula

V₀/V₁ = 1 - (1 / 2fRC)

1 / 2fRC = 1 - (v₀/V₁)

C = 2fR ((1 - (v₀/V₁))⁻¹

Substitute  for,

f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V

C = 2 * 240 * 1 (( 1 - (14/15))⁻¹

C = 62.2μf

The peak-to-peak ripple voltage

= (1 / fRC)V₁

= 1 /  ( (120 * 1 * 62.2) )15V

= 2V

The peak-to-peak ripple voltage = 2V

3 0
3 years ago
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