Answer:
The speed of the 11.5kg block after the collision is V≅4.1 m/s
Explanation:
ma= 4.8 kg
va1= 7.3 m/s
va2= - 2.5 m/s
mb= 11.5 kg
vb1= 0 m/s
vb2= ?
vb2= ( ma*va1 - ma*va2) / mb
vb2= 4.09 m/s ≅ 4.1 m/s
Answer:
0.218
Explanation:
Given that
Total vibrations completed by the wave is 43 vibrations
Time taken to complete the vibrations is 33 seconds
Length of the wave is 424 cm = 4.24 m
to solve this problem, we first find the frequency.
Frequency, F = 43 / 33 hz
Frequency, F = 1.3 hz
Also, we find the wave velocity. Which is gotten using the relation,
Wave velocity = 4.24 / 15
Wave velocity = 0.283 m/s
Now, to get our answer, we use the formula.
Frequency * Wavelength = Wave Velocity
Wavelength = Wave Velocity / Frequency
Wavelength = 0.283 / 1.3
Wavelength = 0.218
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>
Use pythagorean's theorem for this, with 7 as a and 5 as b. pythagorean's theorem says that a^2 + b^2 = c^2, so 7^2 * 5^2 = c^2. this gives you 49 + 25 = c^2, so 74 = c^2. c = sqrt 74, which is approximately 8.60 km
