Answer:
μ = 0.6
Explanation:
given,
speed of car = 29.7 m/s
Radius of curve = 50 m
θ = 30.0°
minimum static friction = ?
now,
writing all the forces acting along y-direction
N cos θ - f sinθ = mg
N cos θ -μN sinθ = mg

now, writing the forces acting along x- direction
N sin θ + f cos θ = F_{net}
N cos θ + μN sinθ = F_{net}

taking cos θ from nominator and denominator




now, inserting all the given values

μ = 0.6
Answer:
-62.45m/s and +62.45m/s
Explanation:
The formula for relativistic speed
This is the speed of A with respect to B

where
will be the velocity of person 1: 39m/s
will be the velocity of person 2: -31m/s (negative because is travelling in opposite direction)
and
the velocity of light: 100m/s
The velocity of person 1 measured by person 2 is:

The velocity of person 2 measured by person 1 is:


The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.
Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire.
The magnetic field generated by wire 1 is:

The magnetic field generated by wire 2 is:

And so, the resultant magnetic field at the point midway between the two wires is
Coffee mug is cylindrical shape. Therefore,
Volume of mug = volume of cylinder = πr²h
r = inner radius of mug. h = height of coffee = 6 cm.
Density of coffee = Density of water = 1 g/ml
Therefore, volume of coffee = mass/density = 400/1 = 400 ml = 400 cm³
Volume of coffee = πr²h
400= πr²(6)
r² = 21.23
r = inner radius of mug = 4.607 cm
.Answer:
The value of the work done is
.
Explanation:
When a charged particle having charge
is moving through an electric field
, the net force (
) on the charge is

and the work done (
) by the particle is

Given,
.
Substitute the value of electric field in equation (1) and then substitute the result in equation (2).
![W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}](https://tex.z-dn.net/?f=W%20%26%3D%26%20%5Cint%5Climits%5E7_0%20%7Bq%5Cdfrac%7BA_%7B0%7D%7D%7Bx%5E%7B1%2F2%7D%7D%7D%20%5C%2C%20dx%20%5C%5C%26%3D%26%20qA_%7B0%7D%20%5Cint%5Climits%5E7_0%20%7Bx%5E%7B-1%2F2%7D%7D%20%5C%2C%20dx%20%5C%5C%26%3D%26%202qA_%7B0%7D%5Bx%5E%7B1%2F2%7D%5D_%7B0%7D%5E%7B7%7D%5C%5C%26%3D%26%205.29%20qA_%7B0%7D)