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2 years ago

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two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire

cted up to the right at 30degrees with the horizontal and force Q is 100 lb directed up to the left at 60degrees with the horizontal, determine the moment of the resultant of these two forces with respect to O.

1 answer:

valentinak56 [21]2 years ago

6 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>

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Television sets. What does it mean to refer to a 40-in. TV set? Such units refer to the diagonal of the screen. A 40-in. TV set

lawyer [7]

Answer:

$h=24\ in$ is the height of the TV screen.

Explanation:

Given:

diagonal of TV screen, $d=40\ in$

width of TV scree, $w=32\ in$

As we know that the TV screen is rectangular in shape so including its height and with along with the diagonal forms a right angled triangle in which the diagonal is the hypotenuse of the triangle formed so.

$h=\sqrt{d^2-w^2}$

$h=\sqrt{40^2-32^2}$

$h=24\ in$ is the height of the TV screen.

3 0

3 years ago

A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of th

SIZIF [17.4K]

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

Volume = 470 cm³

Density = 3.55 g/cm³

Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

Volume = 470 cm³

Density of liquid = 1 g/cm³

$\texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N$

c) Mass of the water displaced = Volume of body x Density of liquid

Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

Weight of rock under water = 16.37 - 4.61 =11.76 N

3 0

3 years ago

Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.

GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately $\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$

With the following condition :

F = gravitational force (N)
G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_1}$ = mass of the first object (kg)
$\sf{m_2}$ = mass of the second object (kg)
r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_X}$ = mass of the planet X = $\sf{1.55 \times 10^{22}}$ kg.
$\sf{m_Y}$ = mass of the planet Y = $\sf{3.95 \times 10^{28}}$ kg.
r = distance between two objects = $\sf{3.75 \times 10^{11}}$ m.

What was asked :

F = gravitational force = ... N

Step by step :

$\sf{F = G \times \frac{m_X \times m_Y}{r^2}}$

$\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}$

$\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}$

$\sf{F \approx 2.9 \times 10^{39 - 22}}$

$\sf{F \approx 2.9 \times 10^{17} \: N}$

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>See More</h3>

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1 year ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

3 years ago

A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charg

iogann1982 [59]

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

$F=\frac{k*q1*q2}{d^{2}}$

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

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3 years ago