At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is
∑ F = ma
n - 430 N = (430 N)/g • a
where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is
a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²
and so
n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N
Answer:
B
Explanation:
Absolute advantage is the capability of a commercial entity to produce goods using fewer resources compared to rivals. Using the same inputs, an entity with an absolute advantage produces a larger output compared to competitors. It means the firm has a lower marginal cost of production. Therefore, its products will have the lowest prices in the market.
Answer:
The block will not move.
Explanation:
We'll begin by calculating the frictional force. This can be obtained as follow:
Coefficient of friction (µ) = 0.6
Mass of block (m) = 3 Kg
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (R) = mg = 3 × 10 = 30 N
Frictional force (Fբ) =?
Fբ = µR
Fբ = 0.6 × 30
Fբ = 18 N
From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.
In 2Cr2O7 there’s 2 items; Cr and O
The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.
<h3>De Broglie wavelength:</h3>
The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.
λ=h/p
Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.
Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.
Therefore, λ=h/(mv)
λ=(6.63×10⁻³⁴)/(0.56×26)
λ=4.55×10⁻³⁵ m.
The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.
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