L = length of the incline = 75 m
θ = angle of incline = 22 deg
h = height of skier at the top of incline = L Sinθ = (75) Sin22 = 28.1 m
μ = Coefficient of friction = 0.090
N = normal force by the surface of incline
mg Cosθ = Component of weight of skier normal to the surface of incline opposite to normal force N
normal force "N" balances the component of weight opposite to it hence we get
N = mg Cosθ
frictional force acting on the skier is given as
f = μN
f = μmg Cosθ
v = speed of skier at the bottom of incline
Using conservation of energy
potential energy at the top of incline = kinetic energy at the bottom + work done by frictional force
mgh = f L + (0.5) m v²
mgh = μmg Cosθ L + (0.5) m v²
gh = μg Cosθ L + (0.5) v²
(9.8 x 28.1) = (0.09 x 9.8 x 75) Cos22 + (0.5) v²
v = 20.7 m/s
On a large scale, no. Sound travels through space by the compression and relaxation of molecules. When you speak to someone on Earth, each inflection of voice compresses local molecules, and relaxes the molecules around the compressed ones. In space, however, the large lack of molecules means that the compression waves can’t be transmitted from the sound to the listener.
Answer:
(a) 0.17 m
(b) 5.003 m
(c) 6.38 × 
N
(d) 7.37 ×
N
Explanation:
(a) The minimum value of 
will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.
(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.
<em>Hence, the maximum distance is 5.002 m</em>
(c) For minimum magnitude we use the minimum distance calculated in (a)
Minimum Distance = 0.17 m
For electrostatic force= 
×
(d) For maximum magnitude, we use the maximum distance calculated in (b)
Maximum Distance = 5.002 m
Using the formula for electrostatic force again:
F = 
F= 7.37×
N
C light energy
The solar energy from the sun converts to chemical potential energy
I'm not from that school but I can help you.
