-- It takes 100 calories of heat to make 10 grams of the stuff 20° warmer.
How much of the heat warms each gram ?
-- It takes 10 calories of heat to make each gram of the stuff 20° warmer.
How much of the heat warms that gram each degree ?
-- It takes 1/2 calorie of heat to make each gram of the stuff 1° warmer.
The specific heat of that stuff is
(1/2 calorie) per gram per °C.
That's choice-3 .
Answer:
High energy waves have high amplitudes
Explanation:
The sound is perceived as louder if the amplitude increases, and softer if the amplitude decreases. ... The amplitude of a wave is related to the amount of energy it carries. A high amplitude wave carries a large amount of energy; a low amplitude wave carries a small amount of energy
The X and Y components are as follows;
1. X = 35 * cos 57 = 19. 1m/s; Y = 35 * sin 57 = 29.4 m/s
2. X = 12 * -cos 34 = -10 m/s; Y = 12 * -sin 34 = -6.7 m/s
3. X = 8 * -cos 90 = 0 m/s; Y = 12 -sin 90 = -8 m/s
4. X = 20 * cos 75 = 5. 2m/s; Y = 20 * (-sin 75) = -19.3 m/s
<h3>What are the horizontal and vertical components of the vectors?</h3>
The horizontal and vertical components of the velocities are given as follows:
- Horizontal component, X = x cos θ
- Vertical component, Y = y sin θ
1. 35 m/s at 57° from x-axis
X = 35 * cos 57 = 19. 1m/s
Y = 35 * sin 57 = 29.4 m/s
2. 12m/s at 34° S of W
X = 12 * -cos 34 = -10 m/s
Y = 12 * -sin 34 = -6.7 m/s
3. 8 m/s at South
X = 8 * -cos 90 = 0 m/s
Y = 12 -sin 90 = -8 m/s
4. 20 m/s at 275° from x-axis
X = 20 * cos 75 = 5. 2m/s
Y = 20 * (-sin 75) = -19.3 m/s
In conclusion, the X and Y components are found by taking cosines and sine of the angles.
Learn more about horizontal and vertical components at: brainly.com/question/26446720
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The charge accumulated in 3.25 μF capacitor is 178.75 μC.
Answer:
Explanation:
In parallel connection, the voltage drop across any passive devices like capacitor or resistor will be constant. So the current flow will be varying in case of parallel connections of capacitors or resistors.
As the capacitance is the measure of amount of charge or current generated for a given amount of voltage, it is directly proportional to the charge or current and inversely proportional to voltage.
C = Q/V
Here the charge accumulated in a capacitor of capacitance 3.25 microfarad need to be determined which is in parallel connection with another capacitor. So the voltage through both the capacitor will be equal to the voltage of the battery which is stated as 55 V.
3.25×
= Q/55
Q = 3.25 * 55 = 178.75 μC
So the charge accumulated in 3.25 μF capacitor is 178.75 μC.
The boiling pot of water is completely open
