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Sunny_sXe [5.5K]

3 years ago

10

A box of mass 21 kg sits on an inclined surface with an angle of 10°. What is

2 answers:

N76 [4]3 years ago

7 0

36 N is the correct answer

alexandr1967 [171]3 years ago

4 0

The component of the weight of the box along the surface is 34.9N.

<u>Explanation:</u>

Given that

Mass of the box m=21kg

Angle of inclination θ=10°

Acceleration due to gravity g=9.8 $m/s^{2}$

Weight of an object is the force with which the earth attracts the object. Since weight is a force its unit is Newton(N). When an object is placed on a surface that has no inclination, its weight is given by the expression

W=mg

where m denotes mass of the object and g denotes the acceleration due to gravity.

When the surface has an inclination, the weight of the object can be resolved into two components that are parallel and perpendicular to the surface.

The component of weight parallel to the surface is given by

W=mgsinθ

where θ is the angle of inclination

The component of weight perpendicular to the surface is given by

W=mgcosθ

In the given problem, we have to find out the parallel component of weight

W=mgsinθ

$W=21\times 9.8 \timessin10\°\\=21\times9.8\times0.17\\=34.9N$

The component of the weight of the box along the surface is 34.9N

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Answer:

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A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force

Oliga [24]

Answer:

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Explanation:

The data for this exercise are as follows:

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4 years ago

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inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

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$-m+2n=2$

Adding both equations:

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Solving:

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Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

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3 years ago