Large bodies of water<span> such as oceans, seas, and large lakes </span>affect<span> the </span>climate<span> of an area. </span>Water<span> heats and cools more slowly than land. Thus, in the summer, the </span>coastal<span> regions </span>will<span> stay cooler and in winter warmer. A more moderate </span>climate<span> with a smaller temperature range </span>is<span> created.</span>
Answer:
1.5 m/s²
Explanation:
For the block to move, it must first overcome the static friction.
Fs = N μs
Fs = (45 N) (0.42)
Fs = 18.9 N
This is less than the 36 N applied, so the block will move. Since the block is moving, kinetic friction takes over. To find the block's acceleration, use Newton's second law:
∑F = ma
F − N μk = ma
36 N − (45 N) (0.65) = (45 N / 9.8 m/s²) a
6.75 N = 4.59 kg a
a = 1.47 m/s²
Rounded to two significant figures, the block's acceleration is 1.5 m/s².
Usually the coefficient of static friction is greater than the coefficient of kinetic friction. You might want to double check the problem statement, just to be sure.
Answer:
the number of grains in the ball is 274,848
Explanation:
Given that;
diameter = 0.5 mm
so radius r = 0.25 mm
first we determine the volume of the ball using the following equation;
V = 4/3×πr³
we substitute
V = 4/3×π(0.25)³
V = 0.06544 mm³
Now form table 1.1 "Grain sizes" a metal with grain size number of 12 has about 4,200,000 grains/mm³
so;
Number of grains N = 0.06544 × 4,200,000
N = 274,848 grains
Therefore, the number of grains in the ball is 274,848
Answer:
The angle for the forward Mach line is 19.47°
The angle for the rearward Mach line is 5.21°
Explanation:
From table A-1 (Modern Compressible Flow: with historical perspective):
(M₁ = 3)
If Po₁ = Po₂

Table A-1:

Table A-5:
v₁ = 49.76°
μ₁ = 19.47°
v₂ = 60.55°
μ₂ = 16°
θ = 60.55 - 49.76 = 10.79°
The angle for the forward Mach line is:
μ₁ = 19.47°
The angle for the rearward Mach line is:
θr = μ₂ - θ = 16 - 10.79 = 5.21°
Answer: 2400m
Explanation: 2400m because 600 times 4 equals 2400