You are worried that a solid substance has dissolved in a beaker of water in a lab. What would be a way to separate the substanc
2 answers:
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Answer:
Answer:
1.1 x 10^9 ohm metre
Explanation:
diameter = 1.5 mm
length, l = 5 cm
Potential difference, V = 9 V
current, i = 230 micro Ampere = 230 x 10^-6 A
radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m
Let the resistivity is ρ.
Area of crossection
A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2
Use Ohm's law to find the value of resistance
V = i x R
9 = 230 x 10^-6 x R
R = 39130.4 ohm
Use the formula for the resistance
ρ = 1.1 x 10^9 ohm metre
Explanation:
1) The coefficient of static friction is 0.980
2) The coefficient of kinetic friction is 0.908
Explanation:
1)
In the first situation, the box is still at rest. There are two forces acting on the box:
- The force of push, F, forward
- The force of static friction,
Since the box is in equilibrium we have
(1)
The value of frictional force changes from zero to a maximum value which is given by:
(2)
where
is the coefficient of static friction
m is the mass of the box
g is the acceleration of gravity
So, if the force F needed to put the box in motion is 48 N, it means that this is also the maximum value of the force of friction. So, we can combine eq.(1) and (2) to find the coefficient of static friction:
where:
F = 48 N
m = 5.0 kg
Substituting,
2)
In this second situation, the object is already in motion, so the equation of motion is:
where
F = 48 N is the force applied forward
is the acceleration of the box
is the force of kinetic friction, where
is the coefficient of kinetic friction
We can therefore rearrange the equation to find the coefficient:
Learn more about friction:
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