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3 years ago

A sphere of diameter 6.0

Physics

1 answer:

LenKa [72]3 years ago

7 0

Answer:

V1 = 4/3 pi R^3 = pi D^3 / 6 D = 2 R volume of sphere

V2 = pi r^2 L = pi d^2 L / 4 volume of wire

V2 / V1 = 1 = 3/2 d^2 L / D^3 since volumes are equal

L = 2/3 D^3 / d^2 = 2/3 * 6^3 / .02^2 = 360,00 cm = 3600 m

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A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a

3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec

Explanation:

We have given acceleration of the car $a_1=5.9m/sec^2$

Acceleration of the stock car $a_2=3.6m/sec^2$

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So $s_1=s_2$

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So $\frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6$

After solving t = 3.56 sec

(b) From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m$

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

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3 years ago

Which excited electron has the lowest energy?

bearhunter [10]

Answer:

The lowest energy of electron is the ground state.

Explanation:

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2 years ago

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How to convert 250 newton to pounds? It is likely to be difficult to move?

maxonik [38]

I have no idea I need the answer too

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2 years ago

Which of the following would add more resistance? Check all that apply.

Blababa [14]

Answer:C. E.

Explanation:

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2 years ago

A curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the

adelina 88 [10]

Answer:

maximum possible velocity = $\sqrt{ugR}$

Explanation:

centripetal acceleration when the car is going in the circle must be less than the maximum friction for the car to not slip.

centripetal acceleration $\frac{mv^{2}}{r}$

where v is the velocity of car and r is the radius of circle

maximum friction = umg

where u is the coefficient of static friction.

therefore $umg\geq \frac{mv^{2}}{R}$

therefore maximum possible velocity = $\sqrt{ugR}$

6 0

3 years ago