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Nikitich [7]
3 years ago
11

Ball Throws

Engineering
1 answer:
professor190 [17]3 years ago
3 0
Jake because the more speed, the more kinetic energy which = force.
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A cylindrical rod of a metal alloy is stressed elastically in tension. The original diameter was 11 mm and a force of 55 kN was
andrey2020 [161]

Answer:

10.984mm

Explanation:

by elastic modulus

stress=modulus of elasticity*strain

stress=loading/area                          area" cross-section"

11mm=0.011m

area=π(d/2)^2=π(0.011/2)^2=9.503*10^-5 square meter

stress=55000/(9.503*10^-5)=578.745 MPa

convert MPa and GPa to pascal.

strain=stress/modulus=(578.745*10^6)/(125*10^9)=0.00463............axial strain

v=Poisson  ratio

lateral strain=(-v)*axial strain= -0.31*0.00463

lateral strain= -1.4353*10^-3=change in diameter/ original diameter

change in diameter=(-1.4353*10^-3)*0.011= -1.57883*10^-5 m

negative indicates decrease in diameter.

decrease in dia.=0.01578mm

new diameter=11-0.01578= 10.984mm

3 0
3 years ago
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str
Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 mm^4

So the shear stress at point  A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

3 0
3 years ago
Some Tiny College staff employees i s are information technology (IT) personnel. Some IT personnel provide technology support fo
Westkost [7]

Answer:

solution in the picture attached

Explanation:

3 0
3 years ago
With which type of media are you interacting when you read an online shopping website? A. film B. print C. electronic D. social
irina [24]

With which type of media are you interacting when you read an online shopping website?

A. film

B. print

C. electronic

<h3><u>D. social </u><u>media</u></h3>

Digital media/social media. Normally through ads

8 0
2 years ago
How much heat is lost through a 3’× 5' single-pane window with a storm that is exposed to a 60°F temperature differential?A. 450
arsen [322]

Answer:

A. 450 btu/h

Explanation:

We solve this problem by using this formula:

Q = U x TD x area

U = U value of used material

TD = Temperature difference = 60°

Q = heat loss

Area = 3x5 = 15

We first find U

R = 1/u

2 = 1/U

U = 1/2 = 0.5

Then when we put these values into the formula above, we would have:

Q = 0.5 x 15 x 60

Q = 450Btu/h

Therefore 450btu/h is the answer

8 0
3 years ago
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