The efficiency of the appliance is 0.62 (62%)
Explanation:
The relationship between the power of the appliance and the energy used by it is
![P=\frac{E}{t}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BE%7D%7Bt%7D)
where
P is the power
E is the energy transferred in input
t is the time elapsed
For this appliance,
P = 7 W
So, in t = 42 s, the energy transferred in input is
![E_{in}=Pt=(7)(42)=294 J](https://tex.z-dn.net/?f=E_%7Bin%7D%3DPt%3D%287%29%2842%29%3D294%20J)
We are told that the useful energy in output however is
![E_{out}=183 J](https://tex.z-dn.net/?f=E_%7Bout%7D%3D183%20J)
Therefore, the efficiency of the appliance is
![\eta=\frac{E_{out}}{E_{in}}=\frac{183}{294}=0.62](https://tex.z-dn.net/?f=%5Ceta%3D%5Cfrac%7BE_%7Bout%7D%7D%7BE_%7Bin%7D%7D%3D%5Cfrac%7B183%7D%7B294%7D%3D0.62)
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F equals ma
F/m equals a
a= 3 then
Answer:
h ’= 12,768 cm
Explanation:
For this exercise let's use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p the distance to the object and q the distance to the image
the magnification equation is
m = h '/ h = -q / p
let's find the distance to the object
1 / p = 1 / f- 1 / q
1 / p = 1/20 - 1 / (- 37.5)
1 / p = 0.076666
p = 13.04 cm
now let's use the magnification equation
h ’= - q / p h
let's calculate
h ’= - (-37.5) / 13.04 4.44
h ’= 12,768 cm
Answer: D. Energy is always conserved
Explanation: The law of energy conservation says that energy is neither created or destroyed, that it is only transferred or transformed from energy. "Conserved" means to maintain a constant overall total, which ties directly into the law of conservation (Energy can't be created nor destroyed, it can only be transferred or transformed from one form to another).
The fraction of Oil will be 0.8and block will be 0.95
What is density?
Density is the ratio of Mass of a compound upon Volume of that compound. It's SI unit can be g/cm³, kg/m³.
As we know the Density of water is 997Kg/m³≈1000Kg /m³.
1) Fraction of Oil is
800/1000=0.8
2) Fraction of Block is
950/1000= 0.95.
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