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 photon travels, on average, a particular distance, d, before being briefly absorbed and released by an atom, which scatters it in a new random direction.
-Given d and the speed of light, c, you can figure out the average time step and space step size (how often the photon “steps” and how far it “steps” each time).
-The size of the Sun is figured in terms of step size. Some surprisingly tricky math happens, involving “Brownian motion” and probabilities. Finally,
-The average time it would take to get to the surface of the Sun is found.
How much heat<span> is </span>required<span> to </span>heat 0.1 g<span> of </span>∆hvap<span> =</span>2260 j/g ∆h<span> =</span>340j/g fus iceat−30 ctosteamat 100c?use<span> the </span>approximate values<span>?</span>
Answer: 1.289 m
Explanation:
The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:
<u>x-component:
</u>
(1)
Where:
is the horizontal distance traveled by the venom
is the venom's initial speed
is the angle
is the time since the venom is spitted until it hits the ground
<u>y-component:
</u>
(2)
Where:
is the initial height of the venom
is the final height of the venom (when it finally hits the ground)
is the acceleration due gravity
Let's begin with (2) to find the time it takes the complete path:
(3)
Rewritting (3):
(4)
This is a quadratic equation (also called equation of the second degree) of the form 
, which can be solved with the following formula:
(5)
Where:
Substituting the known values:
(6)
Solving (6) we find the positive result is:
(7)
Substituting (7) in (1):
(8)
We finally find the horizontal distance traveled by the venom:
The moon's orbital and rotational periods are identical or the same, I<span>ts rate of spin is done in unison with its rate of revolution (the time that is needed to complete one orbit). Thus, the moon rotates exactly once every time it circles the Earth.</span>
