Answer:
113.7
Explanation:
maximum distance (s) = 8.9 km
reference intensity (I0) = 1 x 10^{-12} W/m^{2}
power of a juvenile howler monkey (p) = 63 x 10^{-6} W
distance (r) = 210 m
intensity (I) = power/area
where we assume the area of a sphere due to the uniformity of the output in all directions
area = 4π
= 4π x 
= 554,176.9 m^{2}
intensity (I) = 
therefore the desired ratio I/I0 = 
= 113.7
The horizontal component is calculated as:
Vhorizontal = V · cos(angle)
In your case Vhoriontal = 16 · cos(40) = 12.3 m/s
Answer: 12.3 m/s
Answer:
(i) 208 cm from the pivot
(ii) Move further from the pivot
Explanation:
(i) Sum of the moments about the pivot of the seesaw is zero.
∑τ = Iα
(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0
1250 Nm + 600 N x = 0
x = -2.08 m
Kenny should sit 208 cm on the other side of the pivot.
(ii) To increase the torque, Kenny should move away from the pivot.
<span>First, we use the kinetic energy equation to create a formula:
Ka = 2Kb
1/2(ma*Va^2) = 2(1/2(mb*Vb^2))
The 1/2 of the right gets cancelled by the 2 left of the bracket so:
1/2(ma*Va^2) = mb*Vb^2 (1)
By the definiton of momentum we can say:
ma*Va = mb*Vb
And with some algebra:
Vb = (ma*Va)/mb (2)
Substituting (2) into (1), we have:
1/2(ma*Va^2) = mb*((ma*Va)/mb)^2
Then:
1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2
We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1:
1/2(ma) = (ma^2)/mb
Cancel the ma of the left, leaving the right one with exponent 1:
1/2 = ma/mb
And finally we have that:
mb/2 = ma
mb = 2ma</span>
4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12
