![\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}](https://tex.z-dn.net/?f=%5Cmathfrak%7B%5Chuge%7B%5Corange%7B%5Cunderline%7B%5Cunderline%7BAnSwEr%3A-%7D%7D%7D%7D%7D)
Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
8/4 = y/y-x
8y - 8x = 4y
y = 2x
y = 2 x 4
y = 8
Hope this helps
Answer: An acid is a substance that donates a proton and produces a conjugate base.
Explanation:
According to Bronsted-Lowry theory, an acid is a substance that donates a proton and produces a conjugate base while a base is a molecule or ion which accepts the proton.
An example of Bronsted-Lowry acid and base is Ethanoic acid, CH3COOH and hydroxide ion, OH- respectively as shown in the reaction below
CH3COOH(aq) + OH-(aq) <---> CH3COO-(aq) + H2O(l)
Thus, ethanoic acid acts as an acid by donating a proton to the hydroxide ion which accepts it, thus producing ethanoate ion, CH3COO- as a conjugate base.
You are in an early universe.
In the study of the evolution of the universe, it has been determined before Plank time (before the big bang and right after it), the early universe had the following characteristics:
- There was only one single force acting over all that existed.
- The early universe was very hot and dense because all matter had contracted before the big bang.
- Space and time were wrapped.
These characteristics match the ones described, based on this, we can conclude you are in an early universe.
Learn more about universe in: brainly.com/question/9724831
Answer:
1600 kJ/h per K, 888.88 kJ/h per °F and 888.88kJ/h per R
Explanation:
We make use of relations between temperature scales with respect to degrees celsius:
![1 K= 1^{\circ}C+273\\1^{\circ}F= (1^{\circ}C*1.8)+32\\1 R= (1^{\circ}C*1.8)+491.67](https://tex.z-dn.net/?f=1%20K%3D%201%5E%7B%5Ccirc%7DC%2B273%5C%5C1%5E%7B%5Ccirc%7DF%3D%20%281%5E%7B%5Ccirc%7DC%2A1.8%29%2B32%5C%5C1%20R%3D%20%281%5E%7B%5Ccirc%7DC%2A1.8%29%2B491.67)
This means that a change in one degree celsius is equivalent to a change of one kelvin, while for a degree farenheit and rankine this is equivalent to a change of 1.8 on both scales.
So:
![\frac{Q}{\Delta T(K)}=\frac{Q}{\Delta T(^\circ C)}=1600 \frac{kJ}{h} per K\\\frac{Q}{\Delta T(^\circ F)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per ^\circ F\\\frac{Q}{\Delta T(R)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per R](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7B%5CDelta%20T%28K%29%7D%3D%5Cfrac%7BQ%7D%7B%5CDelta%20T%28%5E%5Ccirc%20C%29%7D%3D1600%20%5Cfrac%7BkJ%7D%7Bh%7D%20per%20K%5C%5C%5Cfrac%7BQ%7D%7B%5CDelta%20T%28%5E%5Ccirc%20F%29%7D%3D%5Cfrac%7BQ%7D%7B%5CDelta%20T%28%5E%5Ccirc%20C%2A1.8%29%7D%3D888.88%20%5Cfrac%7BkJ%7D%7Bh%7D%20per%20%5E%5Ccirc%20F%5C%5C%5Cfrac%7BQ%7D%7B%5CDelta%20T%28R%29%7D%3D%5Cfrac%7BQ%7D%7B%5CDelta%20T%28%5E%5Ccirc%20C%2A1.8%29%7D%3D888.88%20%5Cfrac%7BkJ%7D%7Bh%7D%20per%20R)