Heat is in energy and temperature is measure of energy
The black squirrel has zero kinetic energy (if it's not moving) and lower gravitational potential energy than the red squirrel or zero gravitational potential energy if the ground is assumed to be zero gravitational potential line.
The correct answer is
C. reduce the friction between its moving parts
In fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.
Answer:
The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.
Explanation:
Considering west direction along negative x-axis and north direction along positive y-axis
Given:
The car travels at a speed of 120 km/h in the west direction.
The car then travels at the same speed in the north direction.
Now, considering the given directions, the velocities are given as:
Velocity in west direction is, 
Velocity in north direction is, 
Now, since
are perpendicular to each other, their resultant magnitude is given as:

Plug in the given values and solve for the magnitude of the resultant.This gives,

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.
So, the direction is given as:

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.
Answer:
0.0549 m
Explanation:
Given that
equation y(x,t)=Acos(kx−ωt)
speed v = 8.5 m/s
amplitude A = 5.5*10^−2 m
wavelength λ = 0.5 m
transverse displacement = ?
v = angular frequency / wave number
and
wave number = 2π/ λ
wave number = 2 * 3.142 / 0.5
wave number = 12.568
angular frequency = v k
angular frequency = 8.5 * 12.568
angular frequency = 106.828 rad/sec ~= 107 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)= 5.5*10^−2 cos(12.568 x−107t)
when x =0 and and t = 0
maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))
maximum y(x,t)= 5.5*10^−2 m
and when x = x = 1.52 m and t = 0.150 s
y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )
y(x,t)= 5.5*10^−2 × (0.9986)
y(x,t) = 0.0549 m
so the transverse displacement is 0.0549 m