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vredina [299]
3 years ago
15

Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it

begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere?
Physics
1 answer:
MakcuM [25]3 years ago
5 0

Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface  R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .  

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r =  r-h / r

2h = r-h

3h = r

h = r / 3

You might be interested in
A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 65.5-gram mass is attached at the 21.0-cm
Gnesinka [82]

Answer:

113.53 g

Explanation:

Please see attached photo for explanation.

In the attached photo, M is the mass of the meter stick.

The value of M can be obtained as shown below:

Clockwise moment = M × 10.5

Anticlockwise moment = 65.5 × 18.2

Anticlockwise moment = Clockwise moment

65.5 × 18.2 = M × 10.5

1192.1 = M × 10.5

Divide both side by 10.5

M = 1192.1 / 10.5

M = 113.53 g

Thus, the mass of the meter stick is 113.53 g

7 0
2 years ago
A water tank is in the shape of an inverted cone with depth 10 meters, and top radius 8 meters. Water is flowing into the tank a
Elis [28]

Answer:

The value of leaking rate in the question is repeated. By searching on the web I could find the correct value wich is 0.002h^2 m^3 /min.

The depth of the water has to be equal to 7.07 m in order to have a stationary volume.

Explanation:

In order to have a stationary water level the flow of water that comes into the tank (0.1 m^3/min) must be equal to the flow of water that goes out of the tank (0.002*h^2 m^3/min), therefore:

0.002*h^2 = 0.1

h^2 = 0.1/0.002

h^2 = 50

h = sqrt(50) = 7.07 m

7 0
2 years ago
You’ve had practice calculating the grams of hydrogen gas, but it is also possible to calculate the amount of oxygen gas produce
alekssr [168]

Answer:41.991ml

Explanation:

Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION

2 H+ + 2e → H2 REDUCTION 

Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.

from the oxidation reaction:

1moles of oxygen requires 4moles of electrons to be discharged at the product

F=96500C/mol

Quantity of charge Q=It

=60*60*0.201A

Q=723.6C

Mole=Q/(F*mole ratio of electron)

Mole= 723.6/(4*96500)

Mole=((1809)/(965000))

M=0.0018746114

M1/M2=V1/V2

1/0.00187=22.4dm^3/V2

V2=22.4*0.00187

V2=0.04199129534dm^3

41.99129534ml

5 0
2 years ago
Diffusion and osmosis are forms of passive transport.<br><br> True<br> False
pishuonlain [190]

Answer:

True. Diffusion and osmosis are forms of passive transport.

Explanation:

In diffusion, particles move from an area of higher concentration to one of lower concentration until equilibrium is reached.

In osmosis, a semipermeable membrane is present, so only the solvent molecules are free to move to equalize concentration.

8 0
3 years ago
Read 2 more answers
How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest<br> to 51 m/s?
Nesterboy [21]

Answer:

 t  = 16.5s

Explanation:

Given parameters:

Acceleration = 3.1m/s²

Initial velocity  = 0m/s

Final velocity  = 51m/s

Unknown:

Time taken  = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

        Acceleration  = \frac{v  - u }{t}  

v is the final velocity

u is the initial velocity

t is the time taken

  So;

       3.1  = \frac{51 - 0}{t}  

    3.1t  = 51

       t  = 16.5s

8 0
3 years ago
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