Answer:
30.0625 W
Explanation:
325 g/h x (1h x 1kg)/(3600s x 1000g) x 3,33 x 10^5 J/Kg = 30.0625 J/Kg = 30.0625 W
1) See attached graph
To solve this part of the problem, we have to keep in mind the relationship between current and charge:
where
i is the current
Q is the charge
t is the time
The equation then means that the current is the rate of change of charge over time.
Therefore, if we plot a graph of the charge vs time (as it is done here), the current at any time will be equal to the slope of the charge vs time graph.
Here we have:
- Between t = 0 and t = 2 s, the slope is , so the current is 25 A
- Between t = 2 s and t = 6 s, the slope is , so the current is -25 A
- Between t = 6 s and t = 8 s, the slope is , so the current is 25 A
Plotting on a graph, we find the graph in attachment.
2)
The relationship we have written before
Can be rewritten as
This is valid for a constant current: if the current is not constant, then the total charge is simply equal to the area under the current vs time graph.
Therefore, we need to find the area under the graph.
Here we have a trapezium, where the two bases are
A = 1 ms = 0.001 s
B = 2 ms = 0.002 s
And the height is
h = 10 mA = 0.010 A
So, the area is
So, the charge is .
Answer:
10seconds
Explanation:
use the formula a=v final_v inital/time
To solve this, we are going to use the formula for the kinetic energy of an object:
where
is the kinetic energy of the object.
is the mass of the object.
is the speed of the object.
We know form our problem that the mass of the horse is 500 kilograms, so
; we also know that the speed of the horse is 5 meter/second, so
. Lets replace those values in our formula to find
:
J
We can conclude that the kinetic energy of the horse is
6250 Joules.