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kifflom [539]
3 years ago
7

A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.

Physics
1 answer:
Hoochie [10]3 years ago
8 0

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

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The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5x104 N/C. Determin
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Answer:

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Explanation:

Parameter given:

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(a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.

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F = -1.602 * 10^{-19} * 2.5 * 10^4\\\\\\F = -4.01 * 10^{-15} N

(b) This electrostatic force causes the electron to accelerate with an equivalent force:

F = -ma

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Therefore:

-ma =  -4.01 * 10^{-15} N\\\\\\a = \frac{-4.01 * 10^{-15}}{-m}

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3 years ago
WGVU-AM is a radio station that serves the Grand Rapids, Michigan area. The main broadcast frequency is 1480kHz. At a certain di
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Answer:

a

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b

  w =  9.3 *10^{6} \  rad/s

c

  k = 0.031 m^{-1}

d

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Explanation:

From the question we are told that  

       The frequency of the radio station is  f=  1480 \  kHz  =  1480 *10^{3}\ Hz

         The magnitude of the magnetic field is  B  =  3.0* 10^{-11} \  T

Generally the wavelength is mathematically represented as

          \lambda  =  \frac{c}{f}

Here c is the speed of light with value  c =  3.0 *10^{8} \ m/s

So

        \lambda  =  \frac{3.0 *10^{8}}{ 1480 *10^{3}}

=>      \lambda  =  202.7 \  m

Generally the angular frequency is mathematically represented as

       w =  2 \pi * f

=>   w =  2 * 3.142 *  1480 *10^{3}

=>   w =  9.3 *10^{6} \  rad/s

Generally the wave number is mathematically represented as        

=>      k = \frac{2 \pi }{\lambda}

=>      k = \frac{2  *  3.142  }{ 202.7}

=>      k = 0.031 m^{-1}

Generally the amplitude of the electric field at this distance from the transmitter is mathematically represented as

         E_{max} = c *  B

=>      E_{max} = 3.0 *10^{8} *   3.0* 10^{-11}

=>      E_{max} = 9.0 *10^{-3} \  V/m

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Answer:

The answers and workings is in the Explanation section

Explanation:

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2 resistance in series  R₂ =  6Ω

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The current is 1 Amps

Answer =  6Ω  and 1 Amps

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Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ =  9Ω

The current in the circuit with the three lamps connected in series is

I =V/R₃ =6/9 = 0.67 Amps

The current through the 3 lamps I₃ = 0.67 Amps

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Answer =  $4.32

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