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DENIUS [597]
3 years ago
15

At the same instant that a 0.50-kg ball is dropped from 25m above Earth,? At the same instant that a 0.50-kg ball is dropped fro

m 25m aboveEarth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth’ssurface with an initial speed of 15m/s. They move along nearby lines and pass each other without colliding.At the end of 2.0 s the velocity of the center of mass of the two-ball system is:

Physics
1 answer:
fiasKO [112]3 years ago
6 0
I hope it is clearly visible.. Velocity of the center of mass of 2-ball system is - 11.54m/s. Minus indicates, velocity direction is in downward direction.

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Answer:The human eye is sensitive to yellow-green light having a frequency of about 5.5*10^{14} ... What is the energy in joules of the photons associated with this light? ... As the wavelength and frequency of a wave are related, we can find the energy ... In order to find this value, we need Planck's Constant, h=6.626×10−34 J⋅s h ...

Explanation:

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3 years ago
A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r
Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s²  ,  v = 42 m/s

Explanation:

The uniformly accelerated circular movement,  is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (1)

θ=  ω₀*t + (1/2)*α*t²  Formula (2)

at = α*R  Formula (3)

v= ω*R  Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s  : wheel’s initial angular velocity

R = 1.0 m  : wheel’s radium

(a)  Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ=  ω₀*t + (1/2)*α*t²

θ=  (2)*(10) + (1/2)*(4)*(10)²

θ=  220 rad  

θ=  220 rad  

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

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