Explanation:
Why are high tides found simultaneously on opposite sides of Earth? The tidal bulges occur on both sides of Earth that are aligned with the tide-generating body. The ocean water experiencing high tide rotates around Earth on a 12-hour cycle.
Answer:
They both rises to same height.
Explanation:
When an object is sliding up in friction less surface than according to conservation of energy its potential energy will be converted into kinetic energy.
Here, m is the mass, v is the velocity, g is the acceleration due to gravity and H is the height.
Here the height is independent on the mass of an object and its only depend on velocity.
Now according to the question, two objects have same velocity but they have different masses.
Therefore, they rises to the same height because height will not change with mass.
Answer:
The first frequency of audible sound in the speed sound is
f = 662 Hz
Explanation:
vs = 344 m/s
x = 52 cm * 1 / 100m = 0.52m
The wave length is the distance between the peak and peak so
d = 2x
d = 2*0.52 m
d = 1.04 m
So the frequency in the speed velocity is
f = 1 / T
f = vs / x = 344 m/s / 0.52m
f ≅ 662 Hz
Answer:
Its acceleration is positive
Explanation:
As the car is moving in the negative x-direction than after applying brake then there will be a decrease in the acceleration but in the opposite direction.
As decreasing acceleration consider to be negative but the car is moving in negative direction which means increasing acceleration is negative by sign convention but by applying brake acceleration decrease but in opposite direction than it will give positive value of acceleration.
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>
