Answer:
Energy stored in the capacitor is 
Explanation:
It is given that,
Charge, 
Potential difference, V = 36 V
We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :
U = 0.000027 J
So, the potential energy is stored in the capacitor is . Hence, this is the required solution.
Taiga is the answer.
Hope it helps!
5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5
<h3>What is
displacement ?</h3>
A displacement is a vector in geometry and mechanics that has a length equal to the shortest distance between a point P's initial and final positions. It calculates the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.
The final location xf of a point relative to its beginning position xi, or a relative position (derived from the motion), is another way to express a displacement. The difference between the end and beginning positions can be used to define the equivalent displacement vector
To learn more about displacement from the given link:
brainly.com/question/321442
#SPJ4
Answer:
1.84 kJ (kilojoules)
Explanation:
A specific heat of 0.46 J/g Cº means that it takes 0.46 Joules of energy to raise the temperature of 1 gram of iron by 1 Cº.
If we want to heat 50 g of iron from 20° C to 100° C, we can make the following calculation:
Heat = (specific heat)*(mass)*(temp change)
Heat = (0.46 J/g Cº)*(50g)*(100° C - 20° C)
[Note how the units cancel to yield just Joules]
Heat = 1840 Joules, or 1.84 kJ
[Note that the number is positive: Energy is added to the system. If we used cold iron to cool 50g of 100° C water, the temperature change would be (Final - Initial) or (20° C - 100° C). The number is -1.84 kJ: the negative means heat was removed from the system (the iron).
<h3>
Answer:</h3>
200 kg
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Physics</u>
<u>Newton's Law of Motions
</u>
Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion
Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)
Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction<u>
</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] F = 3000 N
[Given] a = 15 m/s²
[Solve] m = <em>x</em> kg
<u>Step 2: Solve for </u><em><u>m</u></em>
- Substitute in variables [Newton's Second Law of Motion]: 3000 N = m(15 m/s²)
- [Mass] [Division Property of Equality] Isolate <em>m</em> [Cancel out units]: 200 kg = m
- [Mass] Rewrite: m = 200 kg
