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RSB [31]
3 years ago
12

A 4-kW electric heater runs for 2 hours to raise the room temperature to the desired level. Determine the amount of electric ene

rgy used in both kWh and kJ.
Engineering
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

Q' = 8 KW.h

Q'=28800 KJ

Explanation:

Given that

Heat Q= 4 KW

time ,t = 2 hours

The amount of energy used in KWh given as

Q ' = Q x t

Q' = 4 x  2 KW.h

Q' = 8 KW.h

We know that

1 h = 60 min = 60 x 60 s  = 3600 s

We know that W  = 1 J/s

The amount of energy used in KJ given as

Q' = 8 x 3600 = 28800 KJ

Therefore

Q' = 8 KW.h

Q'=28800 KJ

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A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

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6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

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You work for a printing company, and you realize that your colleague sent incorrect price quotes to a client. You begin to write
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The sentence excerpted from the e-mail uses passive voice.  

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