Answer: a) t = 1.8 x 10^2 seconds; b) t = 54 seconds; c) t = 49 seconds. Explanation: a) To determine the time of a stationary passenger on the sidewalk, we use the position formula. Given the constant speed of the walkway, we can calculate the time taken for set distances accordingly. This calculation extends into cases where combined velocities for walking are involved in subsequent queries.
Answer:
The beats frequency measures approximately
4.4 kHz
Explanation:
The beat frequency arises from the original ultrasound frequency, 
, and the frequency of the sound reflected off the car, 
:
(1)
To calculate the frequency of the reflected sound, we apply the Doppler effect formula:
where
v = 340 m/s, the speed of sound
is the velocity of the car
is the frequency of the sound emitted
By substituting values,
Thus, the beat frequency (1) is
The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height.
<span>In this scenario, you can demonstrate it as follows: </span>
<span>the initial velocity is Vo </span>
<span>the launch angle is α </span>
<span>the initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity becomes </span>
<span>Vh = Vo×cos(α) </span>
<span>the total flight duration is the period required to return to a height of 0 m, thus </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 = gravitational acceleration = g (= -9.8 m/s²) </span>
<span>therefore </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 let's examine the horizontal distance. </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>therefore </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 points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set 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>As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
275 kPa Explanation: Here the mass of the gas equals m=1.5 kg with an initial volume of V₁=0.04 m³ and an initial pressure P₁=550 kPa. As provided, the final volume is double the original volume, making V₂ equal to 2 V₁. Since the temperature remains constant, T₁=T₂=T. By substituting the values into the equation... results in final pressure being P₂=275 kPa.
