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3 years ago

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

Physics

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.

Explanation:

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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3 years ago

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Answer:

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3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$