Answer:
3.75 m/s south
Explanation:
Momentum before collision = momentum after collision
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
Since the car and truck stick together, v₁ = v₂.
m₁ u₁ + m₂ u₂ = (m₁ + m₂) v
Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:
(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v
-22500 kg m/s = 6000 kg v
v = -3.75 m/s
The final velocity is 3.75 m/s to the south.
The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration <em>a</em> is such that
4 m/s = <em>a</em> (2.5 s) → <em>a</em> = (4 m/s) / (2.5 s) = 1.6 m/s²
Then the force applied to the box has a magnitude <em>F</em> such that
<em>F</em> = (10 kg) (1.6 m/s²) = 16 N
Answer:
3.2075*10^16
Explanation:
Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.
This problem uses the relationships among current
I, current density
J, and drift speed
vd. We are given the total of electrons that pass through the wire in
t = 3s and the area
A, so we use the following equation to to find
vd, from
J and the known electron density
n,
so:

<span>The current
I is any motion of charge from one region to another, so this is given by:
</span>

The magnitude of the current density is:

Being:

<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd