Refer to the figure shown below.
g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.
Let μ = the coefficient of static friction.
The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N
For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058
The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306
Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306
The electric potential at the origin of the xy coordinate system is negative infinity
<h3>What is the electric field due to the 4.0 μC charge?</h3>
The electric field due to the 4.0 μC charge is E = kq/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q = 4.0 μC = 4.0 × 10 C and
- r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m
<h3>What is the electric field due to the -4.0 μC charge?</h3>
The electric field due to the -4.0 μC charge is E = kq'/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q' = -4.0 μC = -4.0 × 10 C and
- r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m
Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is
E" = E + E'
= -2E
= -2kq/r²
<h3>What is the electric potential at the origin?</h3>
So, the electric potential at the origin is V = -∫₂⁰E".dr
= -∫₂⁰-2kq/r².dr
Since E and dr = dx are parallel and r = x, we have
= -∫₂⁰-2kqdxcos0/x²
= 2kq∫₂⁰dx/x²
= 2kq[-1/x]₂⁰
= -2kq[1/x]₂⁰
= -2kq[1/0 - 1/2]
= -2kq[∞ - 1/2]
= -2kq[∞]
= -∞
So, the electric potential at the origin of the xy coordinate system is negative infinity
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Answer:
<em>B) The disturbance of particles in an area.</em>
Explanation:
<em>A wave involves transmission of energy from one place to another by the actual disturbance of the particles of the medium.</em>
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The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.
Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.
The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.
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You could say almost anything.
For example:
phones,
cars,
computers,
clocks,
hydraulics,
bicycles,
the hadron collider,
Planes,
and so on.
