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3 years ago

4. On the way to the castle, Shrek and Donkey have to walk on a rickety bridgeover

Physics

2 answers:

salantis [7]3 years ago

8 0

Over lava!
I’m pretty sure that’s the answer cuz I haven’t seen the movie in a hot minute!

-YOUR WELCOME!! <3

HACTEHA [7]3 years ago

7 0

They have to walk over lava! They were trying to rescue Fiona and then the dragon appears, Donkey’s love.

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Consider 2 steel rods, A and B, B has three times the area and twice the length of A, so young modulus of B will be what factor

AVprozaik [17]

Answer:

$\frac{Y}{Y_o}$ = 2/3

Explanation:

The yuong modulus of a rod is defined as the relationship between the tensile strength and the strain

Y = $\frac{ \frac{F}{A} }{\frac{\Delta L}{L_o} }$

let's use the subscript "o" for rod A

I = $\frac{ \frac{F}{A_o} }{ \frac{\Delta L}{L_o} }$

tells us that rod B has

A = 3 A₀

L = 2 L₀

we substitute

Y = $\frac{ \frac{F}{A} }{ \frac{\Delta L}{L} }$

Y = $\frac{ \frac{F}{3A_o} }{ \frac{\Delta L}{ 2L_o} }$

y = ⅔ $\frac{ \frac{F}{A_o}}{ \frac{\Delta }{L_o} }$

substituting the value of Y₀

Y = ⅔ Y₀

$\frac{Y}{Y_o}$ = 2/3

5 0

3 years ago

A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.

aleksandrvk [35]

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

$m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c}$ Where $m_{c}$ is the calorimeter mass and $n_{s}$ is the number of moles of the samples; $H_{c}$ is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product $m_{c} *Cp$ is the heat capacity, so the balance equation is:

$6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}$

So, the enthalpy of combustion can be calculated:

$H_{c}=-5899.5\frac{kJ}{mol}$

I will be happy to solve any doubt you have.

4 0

3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

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

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago

A force of 45 N is applied tangentially to the rim of a solid disk of radius 0.12 m. The disk rotates about an axis through its

bearhunter [10]

Answer:

Mass of the disk will be 2.976 kg

Explanation:

We have given force F = 45 N

Radius of the disk r = 0.12 m

Angular acceleration $\alpha =140rad/sec^2$

We know that torque $\tau =I\alpha$

And $\tau =Fr$

So $Fr=I\alpha$ , here I is moment of inertia

So $50\times 0.12=I\times 140$

$I=0.0428kgm^2$

We know that moment of inertia $I=\frac{1}{2}mr^2$

So $0.0428=\frac{1}{2}\times m\times 0.12^2$

m = 2.976 kg

6 0

3 years ago

Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0

3 years ago

4. On the way to the castle, Shrek and Donkey have to walk on a rickety bridgeover​

4. On the way to the castle, Shrek and Donkey have to walk on a rickety bridgeover