This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum. So here they are:
Kinetic energy = (1/2) · (mass) · (speed²)
Momentum = (mass) · (speed)
So, now ... We know that
==> mass = 15 kg, and
==> kinetic energy = 30 Joules
Take those pieces of info and pluggum into the formula for kinetic energy:
Kinetic energy = (1/2) · (mass) · (speed²)
30 Joules = (1/2) · (15 kg) · (speed²)
60 Joules = (15 kg) · (speed²)
4 m²/s² = speed²
Speed = 2 m/s
THAT's all you need ! Now you can find momentum:
Momentum = (mass) · (speed)
Momentum = (15 kg) · (2 m/s)
<em>Momentum = 30 kg·m/s</em>
<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum. When I saw this, I wondered whether that's always true. So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>
Answer:
A. The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
Explanation:
The electric force exerted on a charge by an electric field is given by:
where
F is the force
q is the charge
E is the electric field
We see that if the charge is negative, q contains a negative sign, so the force F and the electric field E will have opposite signs (which means they have opposite directions). This is due to the fact that the direction of the lines of an electric field shows the direction of the electric force experienced by a positive charge in that electric field: therefore, a negative charge will experience a force into opposite direction.
1 W = 1 J/s
Watt<span> is the unit of measure for power .
</span><span>
</span>Joule is the unit of measure for the energy and the second is the unit of measure for time.
Answer:
Explanation:
The strength of the electric field produced by a charge Q is given by
where
Q is the charge
r is the distance from the charge
k is the Coulomb's constant
In this problem, the electric field that can be detected by the fish is
and the fish can detect the electric field at a distance of
Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:
