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irina1246 [14]
3 years ago
13

A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of th

e circle. (A) What is the speed of the stopper at the bottom of the circle? (HINT: Use energy conservation principles!) (10.2 m/s) (B) What is the tension in the string when the stopper is at the top of the circle? (0.27 N) (C) What is the tension in the string when the stopper is at the bottom of the circle?
Physics
1 answer:
Andrew [12]3 years ago
3 0

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2  V² = g x 2r + 1/2 v²

 V² = g x 4r +  v²

 V² = 9.8 x 4 +  8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T =   mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T  - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

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Answer:If the object's speed increases.

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If the object's speed increases, then its kinetic energy will increase. If the kinetic energy increases, the change in kinetic energy will be positive.

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Answer:

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A monochromatic light beam is incident on a barium target that has a work function of 2.50 \mathrm{eV} . If a potential differen
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It is given that work function (Ф) of monochromatic light is 2.50 eV.

Einstein photoelectric equation  is given by:

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where K.E(max) is the maximum kinetic energy.

Substituting (1) into (2) , we get

  ev_0=E-\phi\\1.6\times10^{-19} \times1=E-2.50\\E=1.6\times10^{-19}+2.50\\E=2.50eV

As we know that E=\frac{hc}{\lambda}  ....(3)

where Speed of light,c = 3\times10^8 m/s and Planck's constant , h = 6.63\times 10^-^1^9Js = 4.14\times 10^-^1^5 eVs

From equation (3) , we get

\lambda=\frac{hc}{E} \\\\\lambda=\frac{  4.14\times 10^-^1^5 \times 3 \times10^8}{2.50} \\\\\lambda=\frac{1240\times10^-^9}{2.50} \\\\\lambda=496.8\times10^-^9\\\\\lambda=497nm

Learn about more einstein photoelectric equation  here:

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Explanation:

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G is universal gravitational constant

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Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

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So, the orbital period of a dwarf planet is 138.52 years.

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