Answer:
the stove energy went into heating water is 837.2 kJ.
Explanation:
given,
mass of water = 2000 grams
initial temperature = 0° C
Final temperature = 100° C
specific heat of water (c) = 4.186 joule/gram
energy = m c Δ T
= 2000 × 4.186 × (100° - 0°)
= 837200 J
= 837.2 kJ
hence, the stove energy went into heating water is 837.2 kJ.
The answer is 'A', they travel parallel to the direction of motion
Since the rocket’s acceleration is 3.00 m/s^3 * t, its acceleration is increasing at the rate of 3 m/s^3 each second. The equation for its velocity at a specific time is the integral of the acceleration equation.
<span>vf = vi + 1.5 * t^2, vi = 0 </span>
<span>vf = 1.5 * 10^2 = 150 m/s </span>
This is the rocket’s velocity at 10 seconds. The equation for its height at specific time is the integral velocity equation
<span>yf = yi + 0.5 * t^3, yi = 0 </span>
<span>yf = 0.5 * 10^3 = 500 meters </span>
<span>This is the rocket’s height at 10 seconds. </span>
<span>Part B </span>
<span>What is the speed of the rocket when it is 345 m above the surface of the earth? </span>
<span>Express your answer with the appropriate units. </span>
<span>Use the equation above to determine the time. </span>
<span>345 = 0.5 * t^3 </span>
<span>t^3 = 690 </span>
<span>t = 690^⅓ </span>
<span>This is approximately 8.837 seconds. Use the following equation to determine the velocity at this time. </span>
<span>v = 1.5 * t^2 = 1.5 * (690^⅓)^2 </span>
<span>This is approximately 117 m/s. </span>
<span>The graph of height versus time is the graph of a cubic function. The graph of velocity is a parabola. The graph of acceleration versus time is line. The slope of the line is the coefficient of t. This is a very different type of problem. For the acceleration to increase, the force must be increasing. To see what this feels like slowly push the accelerator pedal of a car to the floor. Just don’t do this so long that your car is speeding!!</span>
A)
2revs in 0.08s
so in 1s thats 25revs
therefore thats <u>50π radians</u> in one second
b)
well, ω=2π/T
therefore ω=50π = 157.079rads^-1
and v=rω where r is in meters;
v=0.3x157.079
<u>v=47.123ms^-1</u>
c)
f=1/T
f=1/period for one rotation
1 rotation = 0.08/2 = 0.04
f=1/0.04
<u>f=25Hz</u>
