Conduction could be the way in which heat is transferred through a sausage to get it up to cooked temperature. convection could be the air around the barbecue and the smell of the barbecue being carried on thermal convection currents. hope this is honest enough
Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx =
and filling in:
Δx =
which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
Answer:
(a) -472.305 J
(b) 1 m
Explanation:
(a)
Change in mechanical energy equals change in kinetic energy
Kinetic energy is given by
Initial kinetic energy is 
Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero
Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence
0-472.305 J=-472.305 J
(b)
From fundamental kinematic equation

Where v and u are final and initial velocities respectively, a is acceleration, s is distance
Making s the subject we obtain
but a=\mu g hence

Answer:
a) The maximum height the ball will achieve above the launch point is 0.2 m.
b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.
Explanation:
a)
For the height reached, we use 3rd equation of motion:
2gh = Vf² - Vo²
Here,
Vo = 3.75 m/s
Vf = 0m/s, since ball stops at the highest point
g = -9.8 m/s² (negative sign for upward motion)
h = maximum height reached by ball
therefore, eqn becomes:
2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²
<u>h = 0.2 m</u>
b)
To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:
2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²
(Vo)² = 19.6 m²/s²
Vo = √19.6 m²/s²
<u>Vo = 4.43 m/s</u>
Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)
<u>Vo = 0.174 in/ms</u>
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