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Ludmilka [50]
3 years ago
12

Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig

htweight material of density 600 kg/m3. (Assume that acceleration due to gravity is g = 9.8 m/s2. Round your answer to one decimal place.)
Physics
1 answer:
Nana76 [90]3 years ago
4 0

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = \frac{1}{3}\pi r^2 h

thus,

volume of the cone = \frac{1}{3}\pi 1.2^2\times 4 = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the \frac{1}{4}times the total height

thus,

center of mass lies at,  h' = \frac{1}{4}\times4=1m

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

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ivann1987 [24]

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f} (1)

Where:

I_{T} - Moment of inertia of the turntable, in kilogram-square meters.

r - Distance of the block regarding the center of the turntable, in meters.

m - Mass of the object, in kilograms.

\omega_{o} - Initial angular speed of the turntable, in radians per second.

\omega_{f} - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

I_{T} = \frac{1}{2}\cdot M\cdot r^{2} (2)

Where M is the mass of the turntable, in kilograms.

If we know that \omega_{o} \approx 7.330\,\frac{rad}{s}, M = 1.5\,kg, m = 0.54\,kg and r = 0.1\,m, then the angular speed of the turntable after the event is:

I_{T} = \frac{1}{2}\cdot M\cdot r^{2}

I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}

I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}

\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}

\omega_{T} = 3.004\,\frac{rad}{s} (28.687\,\frac{rev}{min})

The turntable's angular speed after the event is 28.687 revolutions per minute.

3 0
2 years ago
When a ball is launched from the ground at a 45° angle to the horizontal, it falls back to the ground 50 m from the launch point
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Answer:

Explanation:

Given

angle through which ball is launched=45^{\circ}

Range of ball=50 m

Range of projectile is =\frac{u^2sin2\theta }{g}

50=\frac{u^2sin90}{9.8}

u=22.136 m/s

If ball is thrown straight upward

v^2-u^2=2as

0-(22.136)^2=2(-9.8)s

s=\frac{22.136^2}{2\times 9.8}

s=25 m

(b)For Projectile time of flight is

t=\frac{2usin\theta }{g}

t=\frac{2\times 22.136\times sin45}{9.8}

t=3.19 s

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