Answer:
H(t) = 15 -6sin(2.5π(t -0.5))
Step-by-step explanation:
For midline M, amplitude A, period T and time t0 at which the function is decreasing from the midline, the function can be written as ...
H(t) = M -Asin(2π/T(t -t0))
Using the given values of M=15, A=6, T=0.8 and t0 = 0.5, the equation is ...
H(t) = 15 -6sin(2.5π(t -0.5))
Answer:
Volume of prism = 3,240 cm³
Step-by-step explanation:
GIven.
Hexagonal prism.
Side of base(b) = 12cm
Height of prism = 9cm
Height of base (h)= 10cm
Find:
The volume of the prism.
Computation:
Area of base of hexagonal prism = n/2[bh]
Area of base of hexagonal prism = 6/2[(12)(10)]
Area of base of hexagonal prism = 360 cm²
The volume of prism = Area of base of hexagonal prism × Height of prism
The volume of prism = 360 × 9
Volume of prism = 3,240 cm³
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
