Answer:

Explanation:
The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:





Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:


Lastly, the height reached by the two blocks is:





Answer:
678.2 km/h and 80.54° north of east
Explanation:
From the question,
Using pythagoras theorem,
a² = b²+c²..................... Equation 1
Where a = resultant velocity
Given: b = 600 km/h, c = 100 km/h
Substitute these values into equation 1
R² = 600²+100²
R² = 360000+10000
R² = 460000
R = √460000
R = 678.2 km/h.
And the direction is
tanθ = 600/100
tanθ = 6
tanθ = 6
θ = tan⁻¹(6)
θ = 80.54°.
Hence the resultant velocity of the aircraft is 678.2 km/h and 80.54° north of east
Answer:
F = 24 N
Explanation:
In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end
Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity
∑ τ = 0
F l -x W = 0
F = 
let's calculate
F =
4/100 600
F = 24 N
In order to change the colour of the light, you would need to change both the wavelength and frequency of light, as light in the visible spectrum, have their own unique wavelengths that they occur in along with the measure of frequency, which measures the number of cycles or turns a wave can occur from trough to trough or crest to crest within a given period of time.
Answer:
s = 38.7 m
Explanation:
First we calculate the distance covered during uniform motion of reaction time.
s₁ = vt
where,
s₁ = distance covered during uniform motion = ?
v = uniform speed = 11 m/s
t = time = 2.3 s
Therefore,
s₁ = (11 m/s)(2.3 s)
s₁ = 25.3 m
Now, we calculate the distance covered during decelerated motion:
2as₂ = Vf² - Vi²
where,
a = deceleration = -4.5 m/s²
s₂ = distance covered during decelerated motion = ?
Vf = Final Velocity = 0 m/s
Vi = Initial Velocity = 11 m/s
Therefore,
2(-4.5 m/s²)s₂ = (0 m/s)² - (11 m/s)²
s₂ = (-121 m²/s²)/(-9 m/s²)
s₂ = 13.4 m
the total distance will be:
s = s₁ + s₂
s = 25.3 m + 13.4 m
<u>s = 38.7 m</u>