<span>Answer:
So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z.
So we have x2 + 1002 = z2. Now take its derivative in terms of time to get
2x(dx/dt) = 2z(dz/dt)
So at your specific moment z = 200, x = 100âš3 and dx/dt = +8
substituting, that makes dz/dt = 800âš3 / 200 or 4âš3.
Part 2
sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get
cos a (da./dt) = -100/ z2 (dz/dt)
So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or π/6 radians.
Substitute to get
cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3)
âš3 / 2 (da/dt) = -âš3 / 100
da/dt = -1/50 radians</span>
Answer:
147.456077993 Hz
Explanation:
= Frequency of the sonar = 22 kHz
= Velocity of the whale = 4.95 m/s
v = Speed of sound in water = 1482 m/s
The difference in frequency is given by
The difference in frequency is 147.456077993 Hz
Answer:
0.75 m³/s
Explanation:
Applying,
Q = vA.................... Equation 1
Where Q = flow rate of the water, v = velocity of the water, A = area of the pipe.
From the question,
Given: v = 2.5 m/s, A = 0.3 m²
Substitute these values into equation 1
Q = 2.5(0.3)
Q = 0.75 m³/s
Hence the flow rate of water in the pipe is 0.75 m³/s
answer :
D. 6370.92 m
Explanation:
pls refer to the attachment...
Divide 2500 g by 500 cc. The quotient is 5 g/cc ... the density.