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xz_007 [3.2K]
3 years ago
12

Find the work done by a force of 8 pounds acting in the direction of 44° to the horizontal in moving an object 7 feet from (0, 0

) to (7, 0). Round answer to the nearest tenth of a foot-pound.
Physics
1 answer:
iris [78.8K]3 years ago
8 0

Answer:

40.3 ft-lbs

Explanation:

Work = Force•Distance = (Force)(Distance)(cosΘ)

Note what you know and/or don't know. In this case:

Force = 8lbs

Distance = 7ft

Θ = 44°

So with that and the formula of work above, we can say:

Work = (8lbs)(7ft)(cos44°) = 40.28...ft-lbs

When rounded you have 40.3 ft-lbs of work done by the force.

Side note: Just to be clear, the formula Force•Distance is said as Force "dot" Distance. It's a dot product and not just a multiplication sign; which is why the angle is important to know.

Force×Distance is said as Force "cross" Distance, which relates to a different kind of force and has a slightly different formula.

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3 years ago
A factory has 1200 workers of which 720 are male and the rast are female what percent of workers are female​
umka21 [38]

Answer:

Percent of Female Workers = 40%

Explanation:

The percentage of the female workers in the given group of workers can be easily found by the following formula:

Percent\ of\ Female\ Workers = \frac{No.\ of\ Female\ Workers}{Total\ No.\ of\ Workers}\ x\ 100\%

where,

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3 years ago
The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c
Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

\Sigma F = F_{E}-W = 0 (Eq. 1)

Where:

F_{E} - Electrostatic force exerted on human, measured in Newton.

W - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

q\cdot E-m\cdot g = 0

q = \frac{m\cdot g}{E} (Eq. 2)

E - Electric field, measured in Newtons per Coloumb.

m - Mass, measured in kilograms.

g - Gravity acceleration, measured in meters per square second.

q - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that m = 60\,kg, g = 9.807\,\frac{m}{s^{2}} and E = -150\,\frac{N}{C}, the charge that a 60-kg human must have to overcome weight is:

q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }

q = -3.923\,C

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

F = \kappa \cdot \frac{q^{2}}{r^{2}} (Eq. 3)

Where:

\kappa - Electrostatic constant, measured in Newton-square meter per square Coulomb.

q - Electric charge, measured in Coulomb.

r - Distance between two people, measured in meters.

If we know that \kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q = -3.923\,C and r = 100\,m, then the force of repulsion between two people is:

F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]

F = 13.851\times 10^{6}\,N

The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

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