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tiny-mole [99]
3 years ago
12

After a 65 newton weight has fallen freely from rest a vertical distance of 5.3 meters, the kinetic energy of the weight is

Engineering
1 answer:
inysia [295]3 years ago
3 0

Answer:

The kinetic energy of the weight is 344.5 J

Explanation:

Given that:

Force = F = 65 newton

distance = d = 5.3 meters

We have to find change in kinetic energy ΔK.E

Now we know that, initially kinetic energy was 0 So the formula we use will be:

Work done = Change in kinetic energy

Mathematically,

W =  ΔK.E

As we know W = F . d and  ΔK.E = K.E(final) - K.E(initial)

So by putting values:

F . d = K.E(final) - K.E(initial)

F . d = K.E(final)

As  K.E(initial) is 0 so by putting values of F and d

(65)* (5.3) =  K.E(final)

344.5 J =  K.E(final)

So the change in  K.E will also be 344.5 J

i hope it will help you!

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Answer:

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5 0
4 years ago
Disadvantages of bezier curve and b-spline
Gnesinka [82]

Answer:

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3 years ago
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Answer:

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Explanation:

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3 0
3 years ago
The assembly consists of a brass shell (1) fully bonded to a ceramic core (2). The brass shell [E = 93 GPa, α= 15.1 × 10−6/°C] h
marshall27 [118]

Answer:

ΔT = 62.11°C

Explanation:

Given:

- Brass Shell:

       Inner Diameter d_i = 32 mm

       Outer Diameter d_o = 39 mm

       E_b = 93 GPa

       α_b = 15.1*10^-6 / °C

- Ceramic Core:

       Outer Diameter d_o = 32 mm

       E_c = 310 GPa

       α_c = 3.2*10^-6 / °C

- Unstressed @ T = 8°C

- Total Length of the cylinder L = 160 mm

Find:

Determine the largest temperature increase Δ⁢t that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 60 MPa.

Solution:

- Since, α_b > α_c the brass shell is in compression and ceramic core is in tension. The stress in shell is given as б_a:

                              б_b = - 60 MPa

- The force equilibrium can be written as:

                          б_b*A_b + б_c*A_c = 0

Where, б_b is the stress in core

            A_b is the cross sectional area of the shell

            A_c is the cross sectional area of the core

                           б_b*pi*( d_o^2 - d_i^2) / 4  + б_c*pi*( d_i^2) / 4 = 0

                           б_b*( d_o^2 - d_i^2)  + б_c*( d_i^2) = 0

                           б_c = - б_b*( d_o^2 - d_i^2) / ( d_i^2)

Plug in the values:

                           б_c = 60*( 0.039^2 - 0.032^2) / ( 0.032^2)

                           б_c =  29.121 MPa , б_b = - 60 MPa

-  The total strains in both brass shell and ceramic core is given by:

                           ξ_b = α_b*ΔT + б_b / E_b

                           ξ_c = α_c*ΔT + б_c / E_c

- The compatibility relation is:

                           ξ_b = ξ_c

                           α_b*ΔT + б_b / E_b = α_c*ΔT + б_c / E_c

                           ΔT*(α_b - α_c ) = б_c / E_c - б_b / E_b

                           ΔT = [ б_c / E_c - б_b / E_b ] / (α_b - α_c )

Plug in values and solve:

                           ΔT = [ 0.029121 / 310 + 0.06 / 93 ]*10^6 / (15.1 - 3.2 )

                           ΔT = 62.11°C

8 0
3 years ago
Practice Problem: Large-Particle CompositesThe mechanical properties of a metal may be improved by incorporating fine particles
trasher [3.6K]

Answer: (a). Ec(μ) = 165.6 GPa

(b). Ec(∝) = 83.09 GPa

Explanation:

this is quite straightforward, so we will go step by step.

from the data we have that,

Moduli of elasticity of the metal  -(Em) is 60 Gpa

Moduli of elasticity of oxide is  (Ep) is 380 Gpa

volume Vp = 33% = 0.33

(a). To solve the upper bound-modulus of the elasticity is calculate thus;

Ec (μ) = EmVm + EpVp ----------------(1)

where E rep the modulus of elasticity

v rep the volume fraction

c rep the composite

Vm = 100% - Vp

Vm =  100% - 33% = 67%

Vm = 0.67

substituting the valus of Em, Vm, Ep, Vp  from equation (1) we have;

Ec(μ) = (60×0.67) + (380×0.33)

Ec(μ) = 40.2 + 125.4 = 165.6 GPa

Ec(μ) = 165.6 GPa

(b). The lower bound modulus of elasticity can be calculated thus;

Ec(∝) = EmVp / EpVm + EmVp -------------- (2)

substituting values Em,Vm,Ep,Vp.

Ec(∝) = 60×30 / (380×0.67) + (60 ×0.33)

Ec(∝) = 22800 / 254.6 + 19.8 = 83.09 GPa

Ec(∝) = 83.09 GPa

cheers i hope this helps!!!!

6 0
4 years ago
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