Answer:
The first law, also called the law of inertia, was pioneered by Galileo. This was quite a conceptual leap because it was not possible in Galileo's time to observe a moving object without at least some frictional forces dragging against the motion. In fact, for over a thousand years before Galileo, educated individuals believed Aristotle's formulation that, wherever there is motion, there is an external force producing that motion.
The second law, $ f(t)=m\,a(t)$ , actually implies the first law, since when $ f(t)=0$ (no applied force), the acceleration $ a(t)$ is zero, implying a constant velocity $ v(t)$ . (The velocity is simply the integral with respect to time of $ a(t)={\dot v}(t)$ .)
Newton's third law implies conservation of momentum [138]. It can also be seen as following from the second law: When one object ``pushes'' a second object at some (massless) point of contact using an applied force, there must be an equal and opposite force from the second object that cancels the applied force. Otherwise, there would be a nonzero net force on a massless point which, by the second law, would accelerate the point of contact by an infinite amount.
Explanation:
Angular velocity of the rotating tires can be calculated using the formula,
v=ω×r
Here, v is the velocity of the tires = 32 m/s
r is the radius of the tires= 0.42 m
ω is the angular velocity
Substituting the values we get,
32=ω×0.42
ω= 32/0.42 = 76.19 rad/s
= 76.19×
revolution per min
=728 rpm
Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.
Answer:
The radius of the loop is 20.66 km
Explanation:
let the radius of the loop be r
mass of airplane is m
At the top, the pilot experiences two radial forces, which are
1) Gravitational force is mg
2) Centrifugal forces mv²/r out of the center
When the pilot experiences no weight,
then, mg = mv²/r
r = v² / g
= 450² / 9.8
= 20.66 x 10³3
= 20.66 km
Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (
=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ = 
q₂ = 
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀