The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
They size of the wave and the time of a certain wave.
Answer:
Electromagnetic wave are waves formed as a result of the oscillatory activities involving the electric and the magnetic field.
However in an Electromagnetic wave, the electric field and magnetic field carry equal amounts of energy and the magnitude of the electric field is directly proportional to the magnitude of the magnetic field. This direct proportionality gives rise to the speed of light being the constant between the two fields.
When the electric field is doubled then an equal action is to be set for the magnetic field so it doesn’t deviate from its main functions and characteristics.
The correct answer to this qustion is velocity and time
Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
