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Bumek [7]
3 years ago
11

A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down

the ramp?
Physics
1 answer:
Len [333]3 years ago
4 0

Answer: 0m/s²

Explanation:

Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane

According newton's law of motion

Summation of forces along the plane = mass × acceleration

Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

Ff = nR where

n is coefficient of friction = tan(theta)

R is normal reaction = Wcos(theta)

Fm = Wsin(theta)

Substituting in the formula of newton's first law we have;

Fm-Ff = ma

Wsin(theta) - nR = ma

Wsin(theta) - n(Wcos(theta)) = ma... 1

Given

W = 562N, theta = 30°, n = tan30°, m = 56.2kg

Substituting in eqn 1,

562sin30° - tan30°(562cos30°) = 56.2a

281 - 281 = 56.2a

0 = 56.2a

a = 0m/s²

This shows that the trunk is not accelerating

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An electron (restricted to one dimension) is trapped between two rigid walls 1.40 nm apart. The electron's energy is approximate
Bumek [7]

Answer:

a)    n = 9.9       b)      E₁₀ = 19.25 eV

Explanation:

Solving the Scrodinger equation for the electronegative box we get

         Eₙ = (h² / 8m L²2) n²

where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number

 In this case En = 19 eV let us reduce to the SI system

          En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J

          n = √ (In 8 m L² / h²)

let's calculate

          n = √ (8  9.1 10⁻³¹ (1.4 10⁻⁹)²  30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²

          n = √ (98)            n = 9.9

since n must be an integer, we approximate them to 10

b) We substitute for the calculation of energy

        In = (h² / 8mL2² n²

        In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹  (1.4 10⁻⁹)² 10²

        E₁₀ = 3.08 10⁻¹⁸ J

we reduce eV

      E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)

      E₁₀ = 1.925 101 eV

      E₁₀ = 19.25 eV

the result with significant figures is

        E₁₀ = 19.25 eV

3 0
3 years ago
What is the minimum force require to move a 5kg wooden crate on a wooden floor?
kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with <em>µ</em>. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ <em>F</em> = <em>p</em> - <em>f</em> = 0

• net vertical:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

where

<em>p</em> = magnitude of the <u>p</u>ushing force

<em>f</em> = mag. of <u>f</u>riction

<em>n</em> = mag. of the <u>n</u>ormal force

<em>w</em> = <u>w</u>eight of the crate

The second equation gives

<em>n</em> = <em>w</em> = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of <em>µ</em>, so

<em>f</em> = <em>µ</em> (49 N) = 49<em>µ</em> N

To overcome static friction, the push has to exceed this in magnitude, so that

<em>p</em> > 49<em>µ</em> N

For instance, if <em>p</em> = 0.25, then <em>p</em> would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

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Shkiper50 [21]

Answer:

Option 1 is correct.

The current passing through the brighter bulb is larger.

Explanation:

The brightness of the bulb is determined by the power, I²R

And since they all have equal resistances, the only factor different that could result in more or less power is the current, I through the bulb.

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WARRIOR [948]

Gamma rays are the highest energy EM radiation and typically have energies greater than 100 keV, frequencies greater than 1019 Hz, and wavelengths less than 10 picometers.

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What is potential energy called that is associated with objects that can be stretched
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3 years ago
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