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vova2212 [387]
4 years ago
15

The aluminum rod (E_1 = 75 GPa) is reinforced with the firmly bonded steel tube (E_2 = 209 GPa). The diameter of the aluminum ro

d is d = 28 mm and the outside diameter of the steel tube is D = 42 mm. The length of the composite column is L = 729 mm. A force P is applied at the top surface, distributed across both the rod and tube. Determine the normal stress sigma in the aluminum rod when P = 97 kN:

Physics
1 answer:
Luba_88 [7]4 years ago
5 0

Answer:

35.13Mpa

Explanation:

It's shown in the picture attached

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Its like newtons 3rd law that once in motion a outer force has to stop it
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You go rock climbing with a pack that weighs 70 N and you reach a height of 30 m. how much work did you do to lift your pack? If
Vikentia [17]
When you climb, earth exerts gravitational force on pack in downward direction(pointing towards the center of earth).
In order to climb, you need to work against work done by gravity on the pack.
Hence work done by you = work done by gravity on pack 
                                        = Force x displacement = 70 x 30 = 2100 J.
 
So you need to do 2100 joules of work to lift your pack.

Power is the rate of work done.
Therefore power = work done by you/time(in seconds)
                            =     2100/600 =3.5 watts 
8 0
4 years ago
Radio waves travel at 3.00 · 108 m/s. Calculate the wavelength of a radio wave of frequency 900 kHz. (9.00 · 105 Hz.)
Anuta_ua [19.1K]
V: velocity of wave
f: frequency 
L: wavelenght

v = fL => L = v/f => L = (3x10^8)/(900x10^3) => L = 3.33 x 10^2m
7 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

4 0
3 years ago
A refrigerator having a power rating of 350W operates for 12 hours a day. calculate the cost of electrical energy to operate it
rodikova [14]

Answer:

See the answers below.

Explanation:

The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.

Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]

The fuse can be calculated by knowing the amperage.

P=V*I

where:

P = power = 350 [W]

V = voltage = 240 [V]

I = amperage [amp]

Now clearing I from the equation above:

I=P/V\\I=350/240\\I=1.458[amp]

The fuse should be larger than the current of the circuit, i.e. about 2 [amp]

3 0
3 years ago
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