Answer:
Usually the coefficient of friction remains unchanged
Explanation:
The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.
Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.
Answer:
6.88 mA
Explanation:
Given:
Resistance, R = 594 Ω
Capacitance = 1.3 μF
emf, V = 6.53 V
Time, t = 1 time constant
Now,
The initial current, I₀ = 
or
I₀ = 
or
I₀ = 0.0109 A
also,
I = ![I_0[1-e^{-\frac{t}{\tau}}]](https://tex.z-dn.net/?f=I_0%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D%5D)
here,
τ = time constant
e = 2.717
on substituting the respective values, we get
I = ![0.0109[1-e^{-\frac{\tau}{\tau}}]](https://tex.z-dn.net/?f=0.0109%5B1-e%5E%7B-%5Cfrac%7B%5Ctau%7D%7B%5Ctau%7D%7D%5D)
or
I = ![0.0109[1-2.717^{-1}]](https://tex.z-dn.net/?f=0.0109%5B1-2.717%5E%7B-1%7D%5D)
or
I = 0.00688 A
or
I = 6.88 mA
Answer:
Q = 4.52 10¹⁷ J
Explanation:
Thermal energy can be calculated with
Q = m c_{e} ΔT
in this case it indicates that we approximate seawater to pure water with
c_{e} = 4186 J/ kg K
with the density
ρ = m / V
m = ρ V
V = L³
we substitute
m = ρ L³
Q = ρ L3 c_{e} ΔT
calculate
Q = 1000 (3 103) 3 4186 4
Q = 4.52 10¹⁷ J
12 MPH
I DIDNT do the math my brother did hes in colledge so good lucks guys
