Answer:
A. Normal force is always perpendicular to the area of contact between an object and support.
Explanation:
Normal force is defined as the contact force. If there is no contact between the surfaces, they cannot applies a force which is normal on each other. For e.g, the surfaces of a cubical box and the cart cannot applies a force of normal on each other because of no contact.
If, when there is a contact between two surfaces they applies a normal force on each other, and this force is perpendicular to the each other . This normal force is necessary to prevent object to penetrating into other.
Answer:
Net torque, 
Explanation:
It is given that,
Initial angular speed of the blade, 
Final angular speed of the blade, 
Time, t = 18 s
Radius of the disk, r = 0.13 m
Mass of the disk, m = 0.4 kg
We need to find the net torque applied to the blade. We know that in rotational mechanics the net torque acting on an object is equal to the product of moment of inertia and the angular acceleration such that,
The moment of inertia of the disk, 
Negative sign shows that the net torque is acting in the opposite direction of its motion. Hence, this is the required solution.
Well Bob would need to calculate to net force of someone going down a water slide. Since the person is going down the slide, the person will go faster, depending on their mass/weight and the gravitational pull. As phrased in Newton’s Second Law.
Newton’s Second Law:
Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
Answer:
Pitch. The sound an article makes changes relying upon how quick it is vibrating. At the point when an item vibrates rapidly, sharp sounds are heard. Low-pitched sounds come from things that vibrate all the more gradually.
Explanation:
Answer:
Correct answer: Q = 0.247 μC
Explanation:
Since the oil drop floats, this means that the weight of the drop and the Coulomb force are equal.
F = m · g drop's weight
Fc = k q Q / r² Coulomb force
where it was given:
k = 9 · 10⁹ N m²/C² dielectric constant,
q = 2.2 mC = 2.2 · 10⁻³ C drop's charge,
Q - floor's charge,
r = 7.68 m distance between drop and floor
m = 8.3 grams = 8.3 · 10⁻³ kg drop's weight
and g = 10 m/s²
F = Fc ⇒ k q Q / r² = m · g ⇒ Q = m · g · r²/ k · q
Q = 8.3 · 10⁻³ · 10 · 7.68² / 9 · 10⁹ · 2.2 · 10⁻³
Q = 24.72 · 10⁻⁸ C = 0.2472 · 10⁻⁶ C = 0.2472 μC
Q = 0.247 μC
God is with you!!!
