Imma guess A! Idk if it’s 100% correct tho so I’d check that!
Answer:
Volume = 1,015 acre-feet (Approx)
Explanation:
Given:
Rain = 1.7 in
Time = 30 min
Area = 29 km²
Find:
Volume in acre-feet
Computation:
1 km = 1,000 m
1 m = 3.28 feet
1 km² = 247.105 acre
d = 1.7 in = 1.7 / 12 = 0.14167 ft
Area = 29 × 247.105 = 7,166.045 acre
Volume = 7,166.045 acre × 0.14167 ft
Volume = 1,015 acre-feet (Approx)
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.
Answer:
The nodes and anti nodes would reverse roles.
Explanation:
I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).
Now make these waves out of phase. Then half of the waves will travel half a wavelength farther than the rest. So the path difference will be 0.5λ, 1.5λ, 2.5λ, ....
The net force acting on the object perpendicular to the table is
∑ F[perp] = F[normal] - mg = 0
where mg is the weight of the object. Then
F[normal] = mg = (15 kg) (9.8 m/s²) = 147 N
The maximum magnitude of static friction is then
0.40 F[normal] = 58.8 N
which means the applied 40 N force is not enough to make the object start to move. So the object has zero acceleration and does not move.
