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Marta_Voda [28]

2 years ago

9

The inner surface of a hollow cylinder is subjected to tangential and axial stresses of 40,000 and 24,000 psi, respectively. Det

ermine the maximum shear stress at the inner surface, if the cylinder is pressurized to 10,000 p

1 answer:

Furkat [3]2 years ago

5 0

Answer:

15,000 psi

Explanation:

The solution / solving is attach below.

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A person is planning a bungee jump from a 40 meter high bridge. Under the bridge is a river with crocodiles, so the person does

Nonamiya [84]

Answer:

a. l = 19.7m, b. 18.55m, c. Impact Force = 3889.84 N

Explanation:

The total energy of the system when the person is at top of the bridge is

Potential energy = mgh, Kinetic energy = 0

The total energy of the system when the person reaches just above the surface

Potential energy = 0, Kinetic energy = 0, Spring energy = ½ K X2, where k is the spring constant and X is the deflection

Applying conservation of energy

mgh = 0 + 0 + ½ K X²

80 x 9.81 x 40 = ½ (3600/l) X²

31392 = ½ (3600/l) X²

We can also conclude that

l+ X + 1.75 = 40

l + X = 38.25

a. <u>Substitute the value of x from above into the energy conversion expression</u>

31392 = ½ (3600/l)(38.25 - l)²

31392 x 2/3600 = (38.25 + l² – 2l(38.25))/l

17.44l = l2 – 76.5l + 38.25²

l² – 76.5l – 17.44l +1463.0625 = 0

Solving for l we get

L = 19.7

Hence, length of the rope is 19.7m

b. <u>The deflection is calculated by using the relation between l and X</u>

L + X = 38.25

X = 38.25 – 19.7 = 18.55m

c. <u>The impact force is calculated using the impact force formula which relates the impact force with the deflection</u>

F = KX

F = (3600/l) . X

F = (3600/19.7) . (18.55) = 3889.84 N

Thus, the impact force is 3889.84 N

3 0

2 years ago

a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What

Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, $dq/dt = 100 W$

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, $T_0 = 20^0 C = 273 + 20 = 293 K$

Temperature of the inside of the box, $T_{in} = ?$

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

$dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C$

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3 years ago

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Reptile [31]

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mr_godi [17]

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2 years ago

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bezimeni [28]

Answer:

b. Discharging; anode; cathode

Explanation:

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