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2 years ago

10

consider a solid sphere and a solid disk wiht the same radius and the same mass. explain why the solid disk has a greater moment

of inertia than the solid sphere g

1 answer:

andriy [413]2 years ago

4 0

Answer:

Moment of inertia of the solid sphere:

I

s

=

2

5

M

R

2

.

.

.

.

.

.

.

.

.

.

.

(

1

)

Is=25MR2...........(1)

Here, the mass of the sphere is

M

M

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Sam, whose mass is 75 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 160 N and a coefficien

Volgvan

Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = $\mu *m*g$ .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5 N

F = 86.5 N

so

Acceleration in the First 15 second will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

now After 15 second the velocity will be as

v = u + at ..........3

here u is 0

so v will be

V = 1.15 × 15

v = 17.25

and

now we get travels distance S in 15 s

s = u × t + 0.5 × a × t²

here u is 0

so distance s will be

s = 0.5 × a × t²

s = 0.5 × 1.15 × 15²

s = 129.37 m

and

now acceleration acting is

F = $\mu *m*g$

m a = $\mu *m*g$

a = $\mu* g$

a = - 0.98

here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s

here v velocity is 0 and

u initial velocity is 17.25 m/s

put here value

0 - 17.25² = 2 × (-0.98) × s

solve it we get

s = 151.82 m

so

Total height is

Total height = 129.37 m + 151.82 m

Total height = 281.19 m

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A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle

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<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ )</em>

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