Answer:
A. 91 meters north
Explanation:
Take +y to be north.
Given:
v₀ = 13 m/s
a = 0 m/s²
t = 7 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13 m/s) (7 s) + ½ (0 m/s²) (7 s)²
Δy = 91 m
The displacement is 91 m north.
There are four characteristics of a state. population,government,sovereignty,and territory.
Answer:
According to the law of conservation of energy, energy cannot be created or destroyed, although it can be changed from one form to another. KE + PE = constant. A simple example involves a stationary car at the top of a hill. As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy. In the absence of friction, the car should end up at the same height as it started.
This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.
One can also imagine the energy transformation in a pendulum. When the ball is at the top of its swing, all of the pendulum’s energy is potential energy. When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy. The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms
Answer:
500000000 lbft/s
Explanation:
F = Force or weight = 1000000 lbf
s = Displacement = 10000 ft
t = Time taken = 20 seconds
Work done is given by
Power is given by
One Saucer power is 500000000 lbft/s
Answer:
The observed frequency by the pedestrian is 424 Hz.
Explanation:
Given;
frequency of the source, Fs = 400 Hz
speed of the car as it approaches the stationary observer, Vs = 20 m/s
Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.
The observed frequency is calculated as;
![F_s = F_o [\frac{v}{v_s + v} ] \\\\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C)
where;
F₀ is the observed frequency
v is the speed of sound in air = 340 m/s
![F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%5B%5Cfrac%7B340%7D%7B20%20%2B%20340%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%280.9444%29%20%5C%5C%5C%5CF_o%20%3D%20%5Cfrac%7B400%7D%7B0.9444%7D%20%5C%5C%5C%5CF_o%20%3D%20423.55%20%5C%20Hz%20%5C%5C)
F₀ ≅ 424 Hz.
Therefore, the observed frequency by the pedestrian is 424 Hz.
