The complete question is how much heat energy is delivered by the iron in 30 seconds.
Heat is given by IVt, Where I is the current in Amperes, V is the voltage, and t is the time in seconds,
Therefore;
Heat energy = 120 × 10 ×30
= 36000 joules or 36 kJ
Heat energy is measured in joules.
In science, a broad idea that has been repeatedly verified so as to give scientists great confidence that it represents reality is called "a theory".
<u>Explanation:</u>
In science, the interpretation of a feature of the organic world that can be tested in repeat manner and analysed by applying agreed tests validation methods, calculation and observation in according to the scientific method, such process is called as a theory in science.
The difference lie between a theory and a hypothesis. Because hypothesis is an "educated guess". Overall it is either a proposed interpretation of an observed phenomenon, or a logical inference of a possible causal association between several phenomena.
Answer:
(A) 140 j/sec (b) 1.26 K
Explanation:
We have given the heat heat flowing into the refrigerator = 40 J/sec
Work done = 40 W
(a) So the heat discharged from the refrigerator 
(b) Total heat absorbed =140 j/sec 
Let the temperature be 
Heat absorbed per hour =504000 ![[tex]=400\times 10^3\times \Delta T](https://tex.z-dn.net/?f=%5Btex%5D%3D400%5Ctimes%2010%5E3%5Ctimes%20%5CDelta%20T)
So 
Answer:
Explanation:
We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .
rate of volume flow of one stream
= cross sectional area x velocity
= 8.4 x 3.5 x 2.2 = 64.68 m³ /s
rate of volume flow of other stream
= 6.6 x 3.6 x 2.7
= 64.15 m³ /s
rate of volume flow of rive , if d be its depth
= 11.2 x d x 2.8
= 31.36 d
volume flow of river = Total of volume flow rate of two streams
31.36 d = 64.15 + 64.68
31.36 d = 128.83
d = 4.10 m /s .
Answer:
The new Coulomb force is q₁q₂/9πε₀r²
Explanation
The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².
Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.
Since we have air between q₂ and q₃, the coulomb force between them is
F' = q₂q₃/4πε₀d²
= q₂(q₁/κ)/4πε₀(r/2)²
= 4q₂q₁/κ4πε₀r²
= 4/κ(q₂q₁/4πε₀r²)
= 4/9 × (q₂q₁/4πε₀r²)
= q₁q₂/9πε₀r²
So, the new Coulomb force is q₁q₂/9πε₀r²