Answer:
the planes wings are lifting at an angle to gravity so the plane isn't lifting as much against gravity when it banks. some of the wing lift is going into turning the plane. :) so it needs more lift to bank and stay up
Explanation:
Answer:
Approximately 
(given that the magnitude of this charge is 
.)
Explanation:
If a charge of magnitude 
is placed in an electric field of magnitude 
, the magnitude of the electrostatic force on that charge would be 
.
The magnitude of this charge is 
. Apply the unit conversion 
:
.
An electric field of magnitude 
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Energy is transferred between the ground and the atmosphere via conduction.
Since air is a poor conductor, most energy transfer by conduction<span> occurs right at the earth's surface</span>
We have the equation of motion 
, where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement
Here final velocity, v = 40m/s
Initial velocity, u = 0 m/s
Displacement s = 2 m
Substituting 
So the baseball pitcher accelerates at 400m/
to release a ball at 40 m/s.
<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback
So for conservation of momentum,
rho = mv
M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf
For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2
Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>
