Most of the stars in the milky way will end their lives as

a 1 m length of string is wrapped around a solid disk, of mass .25 kg and a radius of .30m, mounted on a frictionless axle. the

Nezavi [6.7K]

Answer:

60 rad/s

Explanation:

∑τ = Iα

Fr = Iα

For a solid disc, I = ½ mr².

Fr = ½ mr² α

α = 2F / (mr)

α = 2 (20 N) / (0.25 kg × 0.30 m)

α = 533.33 rad/s²

The arc length is 1 m, so the angle is:

s = rθ

1 m = 0.30 m θ

θ = 3.33 rad

Use constant acceleration equation to find ω.

ω² = ω₀² + 2αΔθ

ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)

ω = 59.6 rad/s

Rounding to one significant figure, the angular velocity is 60 rad/s.

8 0

3 years ago

An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left. What is the magnitude and d

ale4655 [162]

An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left.

(1) 22 N to the right
(2) 13 N to the left.

magnitude = 22 - 13
magnitude = 9

Direction would be to the right.

So magnitude is 9N direction to the right.

8 0

3 years ago

6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t

spin [16.1K]

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

5 0

3 years ago

Read 2 more answers

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo

LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0

3 years ago

A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear

JulijaS [17]

Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval (−L,0] for some large number L. The voltage V as a function of x on the interval (0,∞) is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element dx′, we have λ=dqdx′.V(x)=1/4πϵ0∫linedq/r=λ/4πϵ0∫−L0dx/x−x′=λ/4πϵ0(ln|x+L|−ln|x|)

5 0

3 years ago