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3 years ago

How has the use of computer and advance in technology changed healthcare industry

Physics

1 answer:

aev [14]3 years ago

8 0

Yes because we often used it and takes long time to spent and forgot to eat on time.

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A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s

yKpoI14uk [10]

initial velocity=0m/s=u
Acceleration=a=4.8m/s^2
Time=t=15s

Final velocity be v

$\\ \sf\longmapsto v=u+at$

$\\ \sf\longmapsto v=0+4.8(15)$

$\\ \sf\longmapsto v=72m/s$

8 0

3 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

4 years ago

A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50

loris [4]

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 4.5 \times 0.13^2$

U = 0.038 J

at x = 0,the only energy will be kinetic .

$\dfrac{1}{2}mv^2=0.038$

$\dfrac{1}{2}\times 0.23 \times v^2=0.038$

v² = 0.3304

v = 0.575 m/s

displacement of the glider

using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$x =v\sqrt{\dfrac{m}{k}}$

$x =3\times \sqrt{\dfrac{0.23}{4.5}}$

x = 0.678 m

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3 years ago

The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lig

yawa3891 [41]

the answer is C. for plato

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3 years ago

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Finger [1]

you can collect water and shine a light though it and record your finings

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3 years ago

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