Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance

Where, d = distance
A = amplitude
Put the value into the formula


Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Answer:
Newton's second law
Explanation:
The relationship between mass and acceleration is described in Newton's Second Law of Motion. His Second Law states that the more mass an object has, more force is necessary for it to accelerate.
Answer:
Explanation:
The direction of the acceleration is in the same direction as the net force causing it. F = ma is actually a vector equation in which f and a are both vectors and m is a scalar constant.
Answer:
The car stops in 7.78s and does not spare the child.
Explanation:
In order to know if the car stops before the distance to the child, you take into account the following equation:
(1)
vo: initial speed of the car = 45km/h
a: deceleration of the car = 2 m/s^2
t: time
xo: initial distance to the child = 25m
x: final distance to the child = 0m
It is necessary that the solution of the equation (1) for time t are real.
You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:


You take the first value t1 because it has physical meaning.
The solution for t is real, then, the car stops in 7.78s and does not spare the child.
a
a
b
b
a
b
a
This will really help you learn a lot.