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2 years ago

How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s

Physics

1 answer:

solong [7]2 years ago

5 0

Answer:

71 rpm

Explanation:

Given that:

Angular momentum (L) = 0.26

Diameter = 25cm = 0.25 cm

Radius, r = (d/2) = 0.125m

Mass = 5.6 kg

Moment of inertia (I) = 2mr² / 5

I = (2 * 5.6 * 0.125^2) / 5

= 0.175

= 0.175 / 5

= 0.035 kgm²

Angular speed (w) ;

w = L / I

w = 0.26 / 0.035

= 7.4285714

= 7.429 rad/s

w = (7.429 * 60/2π)

w = 445.74 / 2π rpm

w = 70.941724

Angular speed = 70.94 rpm

= 71 rpm

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Calculate the mass of an object with a density of 102.5 g/mL and volume of 375 mL.

AveGali [126]

Answer:

38,437.5

Explanation:

Density(d)= 102.5g/ml

Volume (v)=375ml

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3 years ago

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .

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$A=4 \pi r^{2}\\V=\frac{4}{3} \pi r^{3}$

Therefore, the ratio between the surface area and the volume for the sphere will be:

$\frac{A}{V}=\frac{4 \pi r^{2}\\}{\frac{4}{3} \pi r^{3}}=\frac{3}{r}$

Equating the volume to the constant c, we will find the value of $r$ .

$V=c=\frac{4}{3} \pi r^{3}\\r= (\frac{3c}{4\pi} )^{\frac{1}{3} }$

Substituting the value of r in the ration between surface area and volume, we get:

$\frac{A}{V}=\frac{3}{ (\frac{3c}{4\pi} )^{\frac{1}{3} }}$

Calculating the constants, we get:

$\frac{4.83598}{c^{\frac{1}{3} } }$

Hence, the ration between surface area and volume is $\frac{4.83598}{c^{\frac{1}{3} } }$

To learn more about surface area and volume of sphere, refer to:

brainly.com/question/4387241

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3 years ago

Read 2 more answers

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$