Answer:
B) Angular velocity
Explanation:
The equivalent of Newton's second law for the rotational motions can be written as:
where
is the net torque applied to the object
I is the moment of inertia
is the angular acceleration
From the formula we see that when a constant net torque is applied, then the object also has a constant angular acceleration, .
But we also know that
where 
is the angular velocity: so, a constant angular acceleration means that the angular velocity of the object is changing, so the correct answer is
B) Angular velocity
(moment of inertia and center of gravity do not change since they only depend on the mass and the geometry/shape of the object, which do not change)
Answer:
The magnitude of force is 1.86 N and the direction of force is towards the other wire.
Explanation:
Given:
Current flowing through each power line, I = 130 A
Distance between the two power lines, d = 40 cm = 0.4 m
Length of power lines, L = 220 m
The force exerted by the power lines on each other is given by the relation:
Substitute the suitable values in the above equation.
F = 1.86 N
Since the direction of current flowing through the power lines are opposite to each other, so the force is attractive in nature. Hence, the direction of force experienced by the power lines on each other is towards the each other.
One of Kepler's laws is that the orbits of planets are elliptical. It's not a suggestion.
BTW, circles are ellipses too, but so special that their likelihood is close to zero.
Answer:
a) a = 3.06 10¹⁵ m / s
, b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m 
- m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Answer:
largest lead = 3 m
Explanation:
Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.
So, first we need to calculate the velocities of both the anchorman
given data:
Distance = d = 100 m
Time arrival for A = 9.8 s
Time arrival for B = 10.1 s
Velocity of anchorman A = D / Time arrival for A
=100/ 9.8 = 10.2 m/s
Velocity of anchorman B = D / Time arrival for B
=100/10.1 = 9.9 m/s
As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use
d = vt
= 9.9 x 9.8 = 97 m
So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.
largest lead = 100 - 97 = 3 m
So if his lead no more than 3 m anchorman A win the race.
