Answer:
As an object’s temperature increases, the Rate at which it radiates energy increases.
Answer:
5 is the tripoid stand
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A 15.75-g<span> piece of iron absorbs 1086.75 </span>joules<span> of </span>heat<span> energy, and its ... </span>How many joules<span> of </span>heat<span> are </span>needed<span> to raise the temperature of 10.0 </span>g<span> of </span>aluminum<span> from 22°C to 55°C, if the specific </span>heat<span> of </span>aluminum<span> is o.90 J/</span>g<span>”C2 .</span>
Answer:
A+B; 5√5 units, 341.57°
A-B; 5√5 units, 198.43°
B-A; 5√5 units, 18.43°
Explanation:
Given A = 5 units
By vector notation and the axis of A, it is represented as -5j
B = 3 × 5 = 15 units
Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B
(a) A + B = -5j + 15i
A+B = 15i -5j
|A+B| = √(15)²+(5)²
= 5√5 units
∆ = arctan(5/15) = 18.43°
The angle ∆ is generally used in the diagrams
∆= 18.43°
The direction of A+B is 341.57° based in the condition given (see attachment for diagrams
(b) A - B = -5j -15i
A-B = -15i -5j
|A-B|= √(15)²+(-5)²
|A-B| = √125
|A-B| = 5√5 units
The direction is 180+18.43°= 198.43°
See attachment for diagrams
(c) B-A = 15i -( -5j) = 15i + 5j
|B-A| = 5√5 units
The direction is 18.43°
See attachment for diagram
When an object gets heated by a temperature ΔT energy needed, E = mcΔT
Here energy is given E = 2050 J
Mass of object = 150 g
Change in temperature ΔT = 15
= 15 K
a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.
So heat capacity = E/ΔT = 2050/15 = 136.67 J/K
b) We have E = mcΔT
c = 
So object's specific heat = 911.11 J/kgK