1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
IgorLugansk [536]
3 years ago
9

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

the time required after impact for a puck to lose 10% of its initial speed. Assume air is at 15o C and has a dynamic viscosity of 1.75´10-5 N×s/m2 .
Physics
1 answer:
OverLord2011 [107]3 years ago
3 0

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

You might be interested in
A motorist travels a distance of 406 km during a 7 hr period. What was the average speed in (a) km/hr and (b) m/sec?
Sholpan [36]
The motorist travels (a) 58 km/h and (b) ~16.1 m/sec
4 0
3 years ago
Read 2 more answers
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
CAN SOMEONE HELP ME PLEASE ASAP
Lady_Fox [76]
The third choice is correct
8 0
3 years ago
Read 2 more answers
The fourth harmonic on a string fixed at both ends shows
Charra [1.4K]

Answer:

I believe the answer is B) Two wavelengths

8 0
3 years ago
On a flat surface, a moving bicycle has _____ energy than a stationary car. A. less kinetic B. more kinetic C. less potential D.
Marina86 [1]
On a flat surface a moving bicycle has more kinetic energy than a stationary car
3 0
3 years ago
Read 2 more answers
Other questions:
  • A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it
    15·1 answer
  • A 2.0 kilogram cart moving due east at 6.0 meters per second collides with a 3.0 kilogram cart moving due west. the carts stick
    6·1 answer
  • When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi
    7·1 answer
  • Momentum has a _____.<br> a. direction only<br> b. magnitude only<br> c. direction and magnitude
    14·2 answers
  • When an object has a net force of zero, then it is said to be in what?
    14·1 answer
  • Why is styrofoam not a mineral?
    5·2 answers
  • 3
    9·1 answer
  • Most persons are of the belief that the old time hard or solid iron cars are much safer in collisions than the new model softer
    11·1 answer
  • ! ' 30 !! !
    12·1 answer
  • Question If a force of 2N is used to move a box 1 meter, how much work was done to move the box?​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!