To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.
A) Conservation of Energy,


Here,
m = Mass
v = Velocity
k = Spring constant
A = Amplitude
Rearranging to find the Amplitude we have,

Replacing,


(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.
The Period is defined as

Replacing,


Now the velocity is described as,


We have all the values, then replacing,


Answer:
2.25 Ω
Explanation:
Standard equation
V = IR re-arrange to
V /I = R then sub in the values given
9 / 4 = 2.25 Ω
A). Scale.
You CAN measure volume with all the following except a scale.
Answer:
She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.
Step-by-step explanation:
Given:
The angle of projection of the projectile is,
°
Range of the projectile is,
m.
Acceleration due to gravity, 
The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.
The range of projectile is given as:

Plug in all the given values and solve for minimum speed,
.

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.