Answer:
Sonar
Explanation:
Sonar is a technique that involves the use of sounds in viewing substances in a water medium to aid movement or communication. It makes use of the advantage of sound waves traveling faster and farther in water when compared to other types of waves such as light waves.
During World War II, the military employed the use of SONAR in imaging the seafloor by sending pulses of sound waves down through the water and measuring the time it took for the sound to bounce off the seafloor and return to the receiver.
Answer:
3.0 x10^-3 J
Explanation:
The potential energy of a spring is given by PE = (0.5)k*x^2
Where
K: Spring Constant = 60 N/m
x: displacement of the spring from its equilibrium position = 1cm = 0.01m
Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J
Answer:
117.72kW
Explanation:
Given data
Mass m= 50kg
height x = 2m
time taken = 2 minutes= 129 seconds
let us find the work done
WD= force * distance
WD= mgx
WD= 50*9.81*2
WD= 981 Joules
Let us find the power
Power= work * time
Power= 981*120
Power= 117720
Power= 117.72 kW
Hence the power spent is 117.72kW
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.
Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s
Answer: 11.4 s
Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²
Answer: 44.7 m/s²
