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umka21 [38]
2 years ago
10

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

and the car comes to a rest uniformly in a distance of 160 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Physics
1 answer:
Ludmilka [50]2 years ago
7 0

Answer:

Force applied to stop the car = 1,250 N

Explanation:

Given:

Mass of car (M) = 1,000 kg

Initial velocity (U) = 20 m/s

Final velocity (V) = 0 m/s

Distance (S) = 160 m

Find:

Force applied to stop the car.

Computation:

v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N

Force applied to stop the car = 1,250 N

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2 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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