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ale4655 [162]
2 years ago
9

Which response is the least effective when confronted with trying drugs or alcohol? "I'd rather be in control of what I do."

Physics
1 answer:
SVEN [57.7K]2 years ago
6 0

Answer:

The least effective answer would be "It's okay to lose control every once in a while."

Explanation:

This answer would make it seem like you are either already on drugs or you are willing to try them, which i assume in this case, you are not.

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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 long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
2 years ago
An overhang hollow shaft carries a 900 mm diameter pulley, whose centre is 250 mm from the centre of the nearest bearing. The we
ddd [48]

Answer:yu

Explanation:

6 0
2 years ago
ANSWER BELOW QUESTIONS:
Bogdan [553]

Answer:

why should we do , do by your own , no sense

Explanation:

3 0
2 years ago
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The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp
Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

3 0
1 year ago
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