In which does sound travel fastest? Question 8 options: solids liquids gases plasma.
1 answer:
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The force needed to the stop the car is -3.79 N.
Explanation:
The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.
So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be
Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,
Work done = Final kinetic energy - Initial kinetic energy
Work done = Force × Displacement
So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.
Thus, the force needed to the stop the car is -3.79 N.
Answer:
Ex = kq 2x / ∛ (x² + y²)² and Ex = 2008 N / C
Explanation:
a) The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,
The equation for the electric field produced by a point charge is
E = k q / r²
With r the distance between the point charge and the positive test charge
We look for each electric field
Particle 1. Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance
r² = x² + y²
E1 = k q / (x² + y²)
Particle 2. located at x = -24.9 m
r² = x² + y²
E2 = k q / (x² + y²)
We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.
In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find
Let's measure the angle from axis X
cos θ = CA / H = x / (x2 + y2) ½
E1x = E1 cos θ
E2x = = E1 cos θ
The resulting field
Ey = 0
Ex = E1x + E2x 2 E1x
Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)
Ex = kq 2x / ∛ (x² + y²)²
b) For this part we substitute the numerical values
Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂
Ex = 497.15 x / (x² + 0.062) ³/₂
Point where can the value of the electric field x = 38.1 cm = 0.381 m
Ex = 497.15 0.381 / (0.381² + 0.062) ³/₂
Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2
Ex= 189.41 /0.0943
Ex = 2008 N / C
c) E = 1.00 kN / C = 1000 N / C
To solve this part we must find x in the equation
Ex = 497.15 x / (x² + 0.062) ³/₂
Let's use some arithmetic
Ex / 497.15 = x / (x² + 0.062) ³/₂
[Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃
∛[Ex / 497.15]² = (∛x²) / (x² + 0.062) (1)
The roots of this equation are the solution to the problem,
For Ex = 1.00 kN / C = 1000 N / C
[Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312
1.312 = (∛x² ) / (x² + 0.062)
1.312 (x² + 0.062) = ∛x²
1.312 X² - ∛x² + 1.312 0.062 = 0
1.312 X² - ∛x² + 0.0813 = 0
We need used computer
