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labwork [276]
3 years ago
11

The coefficient of static friction between a block and a horizontal floor is 0.4500, while the coefficient of kinetic friction i

s 0.1500. The mass of the block is 4.400 kg. If a horizontal force is slowly increased until it is barely enough to make the block start moving, what is the net force on the block the instant that it starts to slide?
Physics
1 answer:
amid [387]3 years ago
7 0

Answer:

The net force on the block is 19.40 N.

Explanation:

Given that,

Coefficient of static friction = 0.4500

Coefficient of kinetic friction = 0.1500

Mass of block = 4.400 kg

We need to calculate the net force required to make the block slide

F=\mu_{s} mg

Put the value into the formula

F=0.4500\times4.400\times9.8

F=19.40\ N

Hence, The net force on the block is 19.40 N.

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Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i
garik1379 [7]

Answer

given,

wavelength (λ) = 650 nm

angle = 5°

using bragg's law

sin \theta = \dfrac{n \lambda}{d}

d= \dfrac{n \lambda}{sin \theta}

d= \dfrac{1 \times 650 \times 10^{-9}\ m}{sin5^0}

d = 7.46 x 10⁻⁴ cm

number of slits per centimeter

  = \dfrac{1}{d}\\\Rightarrow \dfrac{1}{7.46\times 10^{-4}}\\\Rightarrow 1340 split per centimeter.

b) wavelength of two rays  650 nm and 420 nm

 d = \dfrac{1}{5000}

     d =  2 x 10⁻6 m

    we now,

sin \theta = \dfrac{n \lambda}{d}

for 650 nm

sin \theta = \dfrac{2\times 650\times 10^{-9}}{2\times 10^{-6}}

\theta =sin^{-1}(0.65)

θ = 40.54°

for 450 nm

sin \theta = \dfrac{2\times 450\times 10^{-9}}{2\times 10^{-6}}

\theta =sin^{-1}(0.45)

θ = 24.83°

now, difference

|θ_{650} -θ_{420}| =40.54°-24.83°

|θ_{650} -θ_{420}| =19.71°

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4 years ago
In a population of seals, most of the seals have similar coloring. However, one seal has albinism. This seal is white and is alm
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Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan
ZanzabumX [31]
From my research, the image supports the question. From the graph given, we can construct the equation of the line using the two-point formula. Using the given value of 601 K, we can solve for the missing value of altitude.

y - y1 = [(y2 - y1)/(x2 - x1)](x- x1)
y - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4)

Substituting y = 601 to solve for x:
601 - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4<span>)
</span>x = 83

Therefore, the probe's instruments will fail at 83 kilometers.

5 0
4 years ago
What type of relationship does this graph show?
Schach [20]

Negative because the graph goes down

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4 years ago
If the rotation of a planet of radius 5.32 × 106 m and free-fall acceleration 7.45 m/s 2 increased to the point that the centrip
leonid [27]

Answer:

v = 6295.55 m/s

Explanation:

Given that,

The radius of a planet, r=5.32\times 10^6\ m

The free fall acceleration of the planet, a = 7.45 m/s²

We need to find the tangential speed of a person standing at the equator.

Also, the centripetal acceleration was equal to the gravitational acceleration at the equator.

We know that,

Centri[etal acceleration,

a=\dfrac{v^2}{r}\\\\v=\sqrt{ar}\\\\v=\sqrt{5.32\times10^6\times 7.45}\\\\v=6295.55\ m/s

So, the tangential speed of the person is equal to 6295.55 m/s.

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3 years ago
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