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Firdavs [7]
3 years ago
11

Which tools would be use to find an irregularly shaped object’s mass and volume?

Physics
1 answer:
madreJ [45]3 years ago
3 0
Scales for weight
Any beaker to measure the volume of liquid displaced
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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul
GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that for spherical shell and pure rolling conditions

v = R \omega

I = \frac{2}{3}mR^2

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh

\frac{5}{6}mv^2 = mgh

h = \frac{5v^2}{6g}

h = \frac{5(8^2)}{6(9.81)} = 5.44 m

Part b)

If ball is not rolling and just sliding over the hill then in that case

\frac{1}{2}mv^2 = mgh

h = \frac{v^2}{2g}

h = \frac{8^2}{2(9.81)} = 3.16 m

3 0
3 years ago
A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second
Scilla [17]

the answer i got is 80 i hope this helpss!!!!

7 0
4 years ago
Read 2 more answers
a flashlight operates with a current of 3.0 a and a power of 4.5 W. what is the voltage of the flashlights battery?
Dafna11 [192]

Answer:

V=1.5V

Explanation:

p=I×V

4.5=3×V

v =  \frac{4.5}{3}

v =  \frac{3}{2}

v=1.5

4 0
2 years ago
30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pu
Mnenie [13.5K]

Answer:

84.05

Explanation:

  • F=mg×0.25
  • F=20x9.81×0.25
  • f=49.05N
  • F=35N

F=f+F

F=49.05+35

=84.05

7 0
3 years ago
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