Answer:
it is for ego control and for super ego control
here's some more information:
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Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences
• net vertical force
F [normal] - F [weight] = 0
• net horizontal force
F [friction] = ma = mv²/r
where v is the tangential speed of the car.
It follows that
F [normal] = F [weight] = mg
and when static friction is maximized at the car's maximum speed,
F [friction] = µ F[normal] = 0.402 mg
Solve for v :
0.402 mg = mv²/r ⇒ v = √(0.402 g (93.5 m)) ≈ 19.2 m/s
Answer:
The mass of unknown object is 8.62Kg
Explanation:
To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.
For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,
By definition we know that the Drag force is defined as
Where,
Drag coefficient
Density
A =Cross-sectional Area
V = Velocity
In the other hand we have,
Where,
Mass of sphere
Mass of unknown object
Equating the two equations we have to
Re-arrange for m_2,
Our values are given by,
Replacing in the equation we have,
<em>Therefore the mass of unknown object is 8.62Kg</em>
Answer:
x =4.5 10⁴ m
Explanation:
To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law
m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg
q = -200 nc (1C / 10 9 nC) = -200 10-9 C
Let's calculate the acceleration
F = ma
F = q E
a = qE / m
a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶
a = 1 10² m / s²
Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)
Vf² = Vo² -2 a x
0 = Vo² - 2 a x
x = Vo² / 2a
x = 3000²/ 2100
x =4.5 10⁴ m
This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial
Second part
In this case Newton's second law is applied on the y axis
F -W = 0
F = w = mg
E q = mg
E = mg / q
E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹
E = 9.8 10⁵ C
The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E
