Answer: -31.36 m/s
Explanation:
This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:
(1)
Where:
is the final velocity of the supply bag
is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)
is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)
is the time
Knowing this, let's solve (1):
(2)
Finally:
Note the negative sign is because the direction of the bag is downwards as well.
Because they have different measurements and weight and mass and some measurements are the same
Answer:
0.217 m/s
Explanation:
The protons in the beam passes undeflected when the electric force is equal to the magnetic force:
qE = qvB
where
q is the proton's charge
E is the magnitude of the electric field
v is the speed of the protons
B is the magnitude of the magnetic field
Re-arranging the equation,

And by substituting
E = 0.5 N/C
B = 2.3 T
We find

Answer:
k = 104.46 N/m
Explanation:
Here we can use energy conservation
so we will have
initial gravitational potential energy = final total spring potential energy
as we know that she falls a total distance of 31 m
while the unstretched length of the string is 12 m
so the extension in the string is given as


so we have



efficiency = (useful energy transferred ÷ energy supplied) × 100
It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:
<u>Way #1:</u>
We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.
(useful energy) / (energy supplied) = (175J) / (250J) = <em>70% efficiency</em>
================================
<u>Way #2: </u>
How much of the energy is wasted ? . . . 75J wasted
What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted
30% of the input energy is wasted. That leaves the other <em>70%</em> to be useful energy.