Answer:
C) 20 m/s
Explanation:
Wave: A wave is a disturbance that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself. Examples of wave are, water wave, sound wave, light rays, radio waves. etc.
The velocity of a moving wave is
v = λf ............................ Equation 1
Where v = speed of the wave, λ = wave length, f = frequency of the wave.
Given: f = 2 Hz (two complete cycles in one seconds), λ = 10 meters
Substituting these values into equation 1
v = 2×10
v = 20 m/s.
Thus the speed of the wave = 20 m/s
The right option is C) 20 m/s
The spontaneous emission of radiations from an unstable nuclei is known as natural radioactivity. on the other hand, The process of emission of radiations from naturally occurring isotopes when they are bombarded with sub-atomic particles or high levels of X-rays or gamma rays called artificial radioactivity.
Answer:
.
Explanation:
If the mass of an object is 
and the velocity of that object is 
, the linear momentum of that object would be 
.
Assume that the initial velocity of the mass is positive (
.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be 
.
Thus, the change in the velocity of the mass would be:
.
The change in the linear momentum of the mass would be:
.
Thus, the magnitude of the change of the linear momentum would be 
.
Answer: acceleration a = 25m/s^2
Explanation:
Given that:
The plane travels with constant acceleration
x1 = 241.22 m at t1 = 3.70 s
x2 = 297.60 m at t2 = 4.20 s
x3 = 360.23 m at t3 = 4.70 s.
We need to calculate the velocity in the two time intervals.
Interval 1:
Average Velocity v1 = ∆x/∆t = (x2 - x1)/(t2-t1)
v1 = (297.60-241.22)/(4.20-3.70) = 112.76m/s
Interval 2:
Average Velocity v2 = ∆x/∆t = (x3-x2)/(t3-t2)
v2 = (360.23-297.60)/(4.70-4.20)
v2 = 125.26m/s
Acceleration:
Acceleration a = ∆v/∆t
∆v = v2-v1 = 125.26m/s-112.76m/s = 12.5m/s
∆t = change in average time of the two intervals = (t3-t1)/2 = (4.70-3.70)/2 = 0.5s
a = 12.5/0.5 = 25m/s^2
Explanation:
14 m is the displacement and towards the north
