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SVEN [57.7K]
3 years ago
5

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surfa

ce is 0.40 and the coefficient of kinetic friction is 0.20.
(a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it?
(b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0N to the box and the box is initially at rest?
Physics
1 answer:
Burka [1]3 years ago
8 0

Answer:

(a) = 0 N

(b) = 2.4 N

Explanation:

given

box of banana weight = 40.0 N

coefficient of static friction  μ = 0.40

coefficient of kinetic friction = 0.20

a).  when no horizontal force is applied .

according to Newton 's third law of motion If there is no force applied to the box,so the frictional  force exerted is 0 N

b) magnitude of friction force

box and the box is initially at rest

    friction force =.Static friction coefficient × weight of the box

      friction force = 0.40 × 6

       friction force =  2.4 N

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Answer:

2.655\times 10^{13}\ photons

Explanation:

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Wavelength of light = \lambda = 570 nm

Pressure from radiation

P=2\frac{I}{c}\\\Rightarrow P=2\frac{1388}{3\times 10^8}\\\Rightarrow P=9.253\times 10^{-6}\ W

Energy of a photon

E=\frac{hc}{\lambda}\\\Rightarrow E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{570\times 10^{-9}}\\\Rightarrow E=3.487\times 10^{-19}\ J

Number of photons

n=\frac{P}{E}\\\Rightarrow n=\frac{9.253\times 10^{-6}}{3.487\times 10^{-19}}\\\Rightarrow n=2.655\times 10^{13}\ photons

Number of photons is 2.655\times 10^{13}\ photons

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3 years ago
Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas
Westkost [7]

Answer:

<em>The range is 15.15 m and the time in the air is 1.01 s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

To calculate the time the object takes to hit the ground, we use the equation below:

\displaystyle t=\sqrt{\frac{2h}{g}}

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

\displaystyle d=15\cdot\sqrt{\frac  {2*5}{9.8}}=15*1.01

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The time in the air is:

\displaystyle t=\sqrt{\frac{2*5}{9.8}}

t = 1.01 s

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Location C is 0.02 m from a small sphere which has a charge of 3 nanocoulombs uniformly distributed on its surface. Location D i
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The change in potential along a path from C to D due to a small charged sphere is 900 V.

Given:

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Distance between the sphere and point C, r₁ = 0.02 m

Distance between the sphere and point D, r₂ = 0.06 m

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We know that the electric potential is given as:

V = k Q/r   - (1)

where, V is the electric potential

            k is Coulomb's force constant

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            r is the  separation distance

The electric potential at point C due to charged sphere can be given as:

V₁ = k Q/r₁

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The electric potential at point D due to charged sphere can be given as:

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Now, the change in potential along the path from C to D can be calculated as:

ΔV = V₂ - V₁

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Answer:

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