Newton’s first law is commonly stated as:
An object at rest stays at rest and an object in motion stays in motion.
However, this is missing an important element related to forces. We could expand it by stating:
An object at rest stays at rest and an object in motion stays in motion at a constant speed and direction unless acted upon by an unbalanced force.
By the time Newton came along, the prevailing theory of motion—formulated by Aristotle—was nearly two thousand years old. It stated that if an object is moving, some sort of force is required to keep it moving. Unless that moving thing is being pushed or pulled, it will simply slow down or stop. Right?
This, of course, is not true. In the absence of any forces, no force is required to keep an object moving. An object (such as a ball) tossed in the earth’s atmosphere slows down because of air resistance (a force). An object’s velocity will only remain constant in the absence of any forces or if the forces that act on it cancel each other out, i.e. the net force adds up to zero. This is often referred to as equilibrium. The falling ball will reach a terminal velocity (that stays constant) once the force of air resistance equals the force of gravity.
Hope this help
Let the distance between the towns be d and the speed of the air be s.
distance = speed * time
convert the minutes time into hours.
When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:
d
s−15
=
7
3
return trip is then :
d
s+15
=
7
5
Cross-multiplying both we get the two-variable system:
3d=7∗(s−15)5d=7∗(s+15)
3d=7s−1055d=7s+105
subtract first equation from second equation we get
2d=210d=105km
Substitute the value of d in the above equations for s.
5∗105=7s+1057s=420s=60km/hr
Answer:
aₓ = 0
, ay = -6.8125 m / s²
Explanation:
This is an exercise that we can solve with kinematics equations.
Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.
x axis
vₓ = v₀ₓ = 1.10 m / s
aₓ = 0
y axis
initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration
= v_{oy} -ay t
ay = (v_{oy} -v_{y}) / t
ay = (0 -10.9) / 1.6
ay = -6.8125 m / s²
the sign indicates that the acceleration goes in the negative direction of the y axis
Answer:
reverberation time appropriate to the use and size of the room, adequate balance between direct and reverberant sound, intimacy and good sound diffusion in the room to obtain a uniform sound.
Explanation:The process of ... second method measured the speed of sound propagation by the phase shift.
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>
