Answer:
Twice
Step-by-Step Explanation:
Time between 7:00 PM and 1:00 AM: 6 hours
Distance: 4818km
Since the distance is 4818km, and the time is 6 hours, you divide 4818 by 6.
803.0000015999 km/h.
The average speed is 803 km/h
Which considering the ideal case scenario if the plane starts at 0 reaches the speed of 803 and the end reduces its speed from 803 to 0. This means we have come across the value of 800 at least twice. Hence, the plane was travelling at a speed of 800 km/h at least 2 times.
A) a mouse, to an order of magnitude = 0.1 m ( a tenth of a meter ) That would be a big mouse but the alternatives are 1 meter or one hundredth of a meter... so go with 1/10th
<span>b) Easy = 1 meter </span>
<span>c) two choices 10m or 100 m . Go with 100 m </span>
<span>d) Stretch it out , trunk tip to tail tip - call it 10 m </span>
<span>e) Your choice 100 m or 1000 m..... These are estimates. So long as you are within one order of magnitude you can't really be given wrong. So I'd say 100m</span>
A group of cells together
<h3><u>Answer;</u></h3>
volume = 6.3 × 10^-2 L
<h3><u>Explanation</u>;</h3>
Volume = mass/density
Mass = 0.0565 Kg,
Density = 900 kg/m³
= 0.0565 kg/ 900 kg /m³
= 6.3 × 10^-5 M³
but; 1000 L = 1 m³
Hence, <u>volume = 6.3 × 10^-2 L</u>
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
