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brilliants [131]
3 years ago
11

A chair of weight 95.0N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 3

9.0N directed at an angle of 37.0 degrees below the horizontal and the chair slides along the floor.
1)Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.

Physics
2 answers:
AlladinOne [14]3 years ago
7 0

Answer:

Answer:

118.4 N

Explanation:

weight of chair, mg = 95 N

Push, F = 39 N

Ф = 37 ° below x axis

Let n be the normal force.

So, by using the diagram and resolve the components of Force F.

n = mg + F SinФ

n = 95 + 39 Sin 37°

n = 95 + 39 x 0.6

n = 118.4 N

Explanation:

mariarad [96]3 years ago
7 0

Answer:118.47

Explanation:

Given

force (F)= 39 N acting at an angle of 37 below horizontal

Force can be divided in to sin and cos component

Thus Fsin37 will act in Y direction

Total Force in Y direction is Normal reaction , Fsin37 and weight

N-mg-Fsin37=0

N=95+39\times \frac{3}{5}

N=118.47 N

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The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

<h3>Net electric force on the charges</h3>

The net electric force on the charges is calculated as follows;

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<h3>Distance between the charges</h3>

|r| = \sqrt{(x_2-x_1)^2+ (y_2-y_1)^2} \\\\|r| = \sqrt{(-4--4)^2+ (0--3)^2} \\\\|r| = \sqrt{(0)^2 + (3)^2} \\\\|r| = 3 \ units

F_{net} = \frac{kq_1q_2}{3^2} \\\\F_{net} = \frac{1}{9} (kq_1q_2) , \  N

Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

Learn more about electric force here: brainly.com/question/17692887

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Which wave must have a medium to travel?
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Letter B

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3 years ago
Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer
tekilochka [14]

Answer: 0.01 m

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h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

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3 years ago
A 110-g bullet is fired from a rifle having a barrel 0.636 m long. Choose the origin to be at the location where the bullet begi
astraxan [27]
<span>Data:
mass =  110-g bullet
d = 0.636 m
Force = 13500 + 11000x - 25750x^2, newtons.

a) Work, W

W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =

W =  13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636  =

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b) x= 1.02 m

</span><span><span>W =  13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02

W = 10383.5

c) %

[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.


</span>
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