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lukranit [14]
2 years ago
6

A ray diagram without the produced image is shown.

Physics
1 answer:
Goryan [66]2 years ago
6 0

Answer:

B) inverted and real

Explanation:

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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
OverLord2011 [107]

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

3 0
2 years ago
A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the
frozen [14]

Answer:

8.0\cdot 10^5 N/C

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

E=k\frac{q}{(R+r)^2}

where

k=8.99\cdot 10^9 N m^2C^{-2} is the Coulomb's constant

q=2.0 \mu C=2.0 \cdot 10^{-6}C is the charge on the sphere

R=10 cm = 0.10 m is the radius of the sphere

r=5.0 cm = 0.05 m is the distance from the surface of the sphere

Substituting, we find

E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C

3 0
2 years ago
A quantity y is to be determined from the equation y=(px)/q^2
ki77a [65]

Answer:

heya answer option b

Explanation:

please mark me brainliest

4 0
1 year ago
When the boy crashes his bumper car into the girl's bumper car, the momentum from his car is transferred to hers. What evidence
Darya [45]

Answer:

Where is the text?

Explanation:

If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.

8 0
2 years ago
Read 2 more answers
In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li
Alenkinab [10]

Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

30 bulbs are of power 60 W

So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

Total power of 80 bulbs = 1.8+5 = 6.8 KW

Total time = 3 hour

We know that energy E=power\times time=6.8\times 3=20.4KWH

Now power of each CFL bulb = 25 W

So power of 80 bulbs = 80×25 = 2000 W = 2 KW

Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

6 0
2 years ago
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