Answer:
The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .
Step-by-step explanation:
Given as :
The mass of liquid water = 50 g
The initial temperature =
= 15°c
The final temperature =
= 100°c
The latent heat of vaporization of water = 2260.0 J/g
Let The amount of heat required to raise temperature = Q Joule
Now, From method
Heat = mass × latent heat × change in temperature
Or, Q = m × s × ΔT
or, Q = m × s × (
-
)
So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )
Or, Q = 50 g × 2260.0 J/g × 85°c
∴ Q = 9,605,000 joule
Or, Q = 9,605 × 10³ joule
Or, Q = 9605 kilo joule
Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer
Answer:
<u>The area of the larger triangle is 315 inches²</u>
Step-by-step explanation:
1. Let's review all the information provided for answering the questions properly:
Pre-image triangle has a base of 7 inches and a height of 10 inches
Scale used : Factor of 3
2. What is the area of the larger triangle??
For calculating the area of the larger triangle, we use the scale this way:
Base = 7 inches * 3
21 inches
Height = 10 inches * 3
30 inches
Area of the larger triangle = 1/2 (Base * Height)
Area of the larger triangle = 1/2 (21 * 30)
<u>Area of the larger triangle = 630/2 = 315 inches²</u>
H represents the height of the ball at a given time, symbolised as t.
Thus, we just need to find when h = 0 so that we find when it hits the ground.
0 = -4.9t² + 19.6t + 58.8
0 = 4.9t² - 19.6t - 58.8
0 = 49t² - 196t - 588
0 = t² - 4t - 12
0 = (t - 6)(t + 2)
So, t = 6 or -2, but t ≠ -2, since time cannot be negative in this instance.
Hence, at 6 seconds, the ball will strike the ground.