Amplitude of vibration is the maximum displacement of an object that has been moved from one point to another due to a vibrating body or its wave measured from an equilibrium position. In short, it is the distance of the object that moved from the vibration.
Answer:
h = 1.94 m
Explanation:
When the bull dog and skate board reach the bottom of the well, all of its potential energy is converted to the kinetic energy:
Kinetic Energy Gained by Bull Dog and Skate Board = Potential Energy Lost by Bull Dog and Skate Board
K.E = P.E
K.E = mgh
h = K.E/mg
where,
h = depth of well = ?
K.E = Kinetic Energy at bottom = 380 J
m = mass of bull dog and skate board = 20 kg
g = 9.8 m/s²
Therefore,
h = 380 J/(20 kg)(9.8 m/s²)
<u>h = 1.94 m</u>
Answer:
The first interval is walked slowly, this is a straight line with a small slope
Second interval stops, which gives a horizontal line, indicating the same position
Third interval, walk back, straight downhill
Explanation:
In this problem we have a uniform movement, this means that the acceleration in each intervals
x = v t
The first interval is walked slowly, this is a straight line with a small slope
Second interval stops, which gives a horizontal line, indicating the same position
Third interval, walk back, straight downhill
Infrared radiation extends from the nominal red edge of the visible spectrum at 700 nanometers (nm) to 1 mm. Ultraviolet<span> </span>light<span> has a radiation with a wavelength shorter than a visible </span>light<span>, longer than X-rays, in the range of 10 nm to 400 nm. Hoped this helped you!!</span>
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m