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lianna [129]
3 years ago
8

An Atwood machine consists of two masses, mA = 7.0 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

tate about a fixed axis. The pulley is a solid cylinder of radius R0 = 0.40 m and mass 0.70 kg .
Physics
1 answer:
erica [24]3 years ago
3 0

Answer:

Explanation:

Given Information

Block A mass, mA = 7 Kg

Block B mass, mB=8 Kg

Pulley mass, M=0.7 kg

Pulley Radius, R=0.4 m

I=moment of inertia=MR^2/2 =0.35R^2

Looking for

Acceleration of both blocks.

Percentage error of the two cases (the pulley’s moment of inertia is ignored and the pulley’s moment of inertia is not ignored).

1) mass acceleration a = g(mB – mA)/(mB + mA + I/R^2) = 9.8*1/(8+7+0.35) = 0.638 (m/s2), cylinder acceleration ε = a/R = 1.6m(s-2)

2)

In case the moment of inertia of the pulley is ignored mass acceleration a1 = 9.8*1/(8+7) = 0.653 (m/s2), % error = 100(aignored-a)/aignored = 2.30%.

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It would be B. The World Cup

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Hecto the Mornar a fursa explain i magnituse of the force acting right angle to the moment arm​
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7 0
3 years ago
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o
Ad libitum [116K]

Answer:

The thickness of the oil slick is 1.95\times10^{-7}\ m

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

2nt= m\lambda

Where, n = Index of refraction

t = thickness

\lambda = wavelength

Put the value into the formula

2\times1.28 \times t=1\times\times500\times10^{-9}

t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }

t=1.95\times10^{-7}\ m

Hence, The thickness of the oil slick is 1.95\times10^{-7}\ m

4 0
3 years ago
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
4 years ago
a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(
horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.    

Explanation:

We can find the velocity of the skier by energy conservation:

E_{1} = E_{2}

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}    (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²  

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}  

v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s    

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!  

6 0
3 years ago
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