I THINK C BECAUSE IF IT IS A GLASS BOX HOW DID A CACTUS GET IN AND NOTHING CAN GET IN OR OUT OF THE BOX SO THERE IS NO CACTUS IN THE BOX
Answer:
When riding on a plane, passengers not only move through space but they move through time as well.
Explanation:
Sure !
Start with Newton's second law of motion:
Net Force = (mass) x (acceleration) .
This formula is so useful, and so easy, that you really
should memorize it.
Now, watch:
The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of 2.5 m/s² .
What's the net force on the box ?
Net Force = (mass) x (acceleration)
= (5.25 kilograms) x (2.5 m/s²)
Net force = 13.125 newtons .
But hold up, hee haw, whoa ! Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?
==> Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.
How much ?
(Bella's 15.75 newtons) minus (13.125 that the box feels)
= 2.625 newtons backwards, applied by friction.
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 .