As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.
Then friction force = 80 Newtons but in the opposite direction.
Friction force = Mu * Normal force exerted by ground = Mu * weight of box
So we find Mu.
Mu = coefficient of friction between box and horizontal surface
= Force of friction / weight = 80 / 50 * 9.81 = 0.163
When an identical box is placed on top, the force of friction is
= Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons
The physical law that explains that is the law of conservation of energy which states that he energy of an isolated sistem remains constant
For a standing wave on a string, the wavelength is equal to twice the length of the string:
In our problem, L=50.0 cm=0.50 m, therefore the wavelength of the wave is
And the speed of the wave is given by the product between the frequency and the wavelength of the wave:
Answer:
See Explanation
Explanation:
Sound is a mechanical wave. A mechanical wave requires a material medium for propagation. This means that sound waves must be carried in air. If there are no air molecules, sound waves can not travel.
When air is gradually removed from the jar by the pump, the sound intensity from the bell gradually decreases owing to the fact that air which is the medium through which sound waves are propagated is gradually being removed from the jar.
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>
