C would be the right answer edu
Answer:
the Jack speed is greater than the Nas speed
v₁ = 2.08 m / s > v2 = 1.85 m / s
Explanation:
For this exercise we can find the average speed of each of the boys, the average speed is defined as the displacement in the time interval
v = x / t
Jack.
It tells us that it travels x = 15 km in a time of t = 2 h
let's reduce the magnitudes to the SI system
x = 15 km (1000 m / 1 km) = 15 10³ m
t = 2 h (3600 s / 1 h) = 7200 s
let's calculate
v₁ = 15 10³/7200
v₁ = 2.08 m / s
Nas travels a distance of x = 10 km in a time of t = 1.5 h
x = 10 km = 10 10³ m
t = 1.5 h (3600s / 1h) = 5400 s
let's calculate the speed
v2 = 10 10³/5400
v2 = 1.85 m / s
From these results we can see that the Jack speed is greater than the Nas speed
L = length of the meter stick = 1 m
h = height of center of mass of stick from bottom end on the floor = L/2 = 1/2 = 0.5 m
m = mass of the meter stick
I = moment of inertia of the meter stick about the bottom end
w = angular velocity as it hits the floor
moment of inertia of the meter stick about the bottom end is given as
I = m L²/3
using conservation of energy
rotational kinetic energy of meter stick as it hits the floor = potential energy when it is vertical
(0.5) I w² = m g h
(0.5) (m L²/3) w² = m g h
( L²) w² = 6g h
( 1²) w² = 6 (9.8) (0.5)
w = 5.4 rad/s
a. The force applied would be equal to the frictional
force.
F = us Fn
where, F = applied force = 35 N, us = coeff of static
friction, Fn = normal force = weight
35 N = us * (6 kg * 9.81 m/s^2)
us = 0.595
b. The force applied would now be the sum of the
frictional force and force due to acceleration
F = uk Fn + m a
where, uk = coeff of kinetic friction
35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)
uk = 0.533
Energy of wave depends on its amplitude and it is given as
here k = constant
A = amplitude
so energy will increase or decrease depends on the amplitude of the wave
So here if we need to check which wave has lower energy then we need to compare the amplitude.
If the amplitude is less then energy must be less
So please check in the figure that which wave out of A and B has lesser amplitude to find out the wave of lesser energy
