Answer: Wave speed may equal frequency*wavelength. Yet doubling the frequency only halves the wavelength; wave speed remains the same. To change the wave speed, the medium would have to be changed. 24. What are some simple steps I can take to protect my privacy online? Many people ... So if you double the frequency and keep the speed constant, the wavelength halves to give the same speed with the doubled frequency. 3.8k views ... The period of a note is 0.3 seconds and the speed of sound in air is 340 m/s. So if you double the frequency and keep the speed constant, the wavelength halves to give the same speed with the doubled frequency. What is the period of a wave if the wavelength is 100m and the speed is 200 m/s? ... If you move towards a light source, the wavelength decreases.
Explanation:
Answer:
H = start height (v = 0)
h = present height
v = present speed
assuming no friction
total energy = PE + KE
mgH = mgh + .5mv^2
if PE = KE then
mgH = mgh + mgh
h = H/2
potential energy = kinetic energy when object is at half its start height.
Explanation:
I’m not sure I think it’s A
Newton's 2nd law:
Fnet = ma
Fnet is the net force acting on an object, m is the object's mass, and a is the acceleration.
The electric force on a charged object is given by
Fe = Eq
Fe is the electric force, E is the electric field at the point where the object is, and q is the object's charge.
We can assume, if the only force acting on the proton and electron is the electric force due to the electric field, that for both particles, Fnet = Fe
Fe = Eq
Eq = ma
a = Eq/m
We will also assume that the electric field acting on the proton and electron are the same. The proton and electron also have the same magnitude of charge (1.6×10⁻¹⁹C). What makes the difference in their acceleration is their masses. A quick Google search will provide the following values:
mass of proton = 1.67×10⁻²⁷kg
mass of electron = 9.11×10⁻³¹kg
The acceleration of an object is inversely proportional to its mass, so the electron will experience a greater acceleration than the proton.