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3 years ago

A diver is on a board 1.80 m above

Physics

2 answers:

Virty [35]3 years ago

8 0

Answer:

v = 6.95 m/s

Explanation:

Given that,

A diver is on a board 1.80 m above the water, s = 1.8 m

The initial speed of the diver, u = 3.62 m/s

Let v is the speed with which she hit the water. It will move under the action of gravity. Using the equation of motion as follows :

$v^2-u^2=2gs\\\\v=\sqrt{u^2+2gs} \\\\v=\sqrt{(3.62)^2+2(9.8)(1.8)} \\\\v=6.95\ m/s$

So, she will hit the water with a speed of 6.95 m/s.

love history [14]3 years ago

4 0

Answer:

6.95

Explanation:

Acellus

Correct

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The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

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b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

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For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

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(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

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3 years ago

An object is placed 100 cm in front of a diverging lens of focal length -25cm. A converging lens of focal length 33 1/3 cm is pl

bagirrra123 [75]

We use 1/o + 1/i = 1/f where o is the distance of the object, i as distance of the image and f is the focal length.
Substituting, 1/ 100 + 1 / i = - 1 /25 
i = - 20 cm 

For the case of the problem,

o = (20 + 30) = 50 cm 

f = 33.33. Using 1 / i + 1 / o = 1/f , i = 100 cm 

M = magnification = - i / o 

m1 = -(-20)/100 = 20/100 = 0.2 

m2 = -100/50 = -2 

M = m1*m2 = -2 x 0.2 = -0.4.

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