Answer:
18.5 m/s
Explanation:
On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:
where
is the coefficient of static friction between the tires and the road
m is the mass of the car
g is the gravitational acceleration
v is the speed of the car
r is the radius of the curve
Re-arranging the equation,
And by substituting the data of the problem, we find the speed at which the car begins to skid:
Answer:
B) Friction
Explanation:
The main source of error is the omission of the effect from friction between block and incline, which is directly proportional to the mass of the block. The force of gravity is constant. The friction force dissipates part of the gravitational potential energy, generating a final speed less than calculated under the consideration of a conservative system. Air resistance is neglected at low speeds like this case.
Answer:
2n t = m λ₀
, R = 0.240 mm
Explanation:
The interference by regency in thin films uses two rays mainly the one reflected on the surface and the one reflected on the inside of the film.
The ray that is reflected in the upper part of the film has a phase change of 180º since the ray stops from a medium with a low refractive index to one with a higher regrading index,
-This phase change is the introduction of a λ/2 change
-The ray passing through the film has a change in wavelength due to the refractive index of the medium
λ₀ = λ / n
Therefore Taking into account this fact the destructive interference expression introduces an integer phase change, then the extra distance 2t is
2 t = (m’+ ½ + ½) λ₀ / n
2t = (m’+1) λ₀ / n
m = m’+ 1
2n t = m λ₀
With m = 0, 1, 2, ...
Where t is the thickness of the film, n the refractive index of the medium, λ the wavelength
The thickness of a hair is the thickness of the film t
2R = t
R = t / 2
R = 0480/2
R = 0.240 mm
Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J
Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
