In the first case, the force acting on the spring is the weight of the mass:
This force causes a stretching of
on the spring, so we can use these data to find the spring constant:
In the second case, the first mass is replaced with a second mass, whose weight is
And since we know the spring constant, we can calculate the new elongation of the spring:
Answer:
Magnitude of fourth displacement is approximately 95 metres,
Direction of fourth displacement is straight west.
Complete question is;
Does the galvanometer deflect to the left or the right when
a) the magnet is being pushed in
b) the magnet is being pulled out
c) the magnet is being held steady?
Answer:
Option A - when the magnet is being pulled out
Explanation:
Faraday’s law of electromagnetic induction states that: “Voltage is induced in a circuit whenever relative motion exists between the conductor and the magnetic field, and the magnitude of the voltage will be proportional to the rate of change of the flux”.
Now, applying it to the question, When the magnet is moved towards the sensitive center of the galvanometer and then pulled out, the needle of the galvanometer will deflect away from its center position in one direction only but when it is held steady, the needle of the galvanometer will return back to zero.
Answer:
#_photons = 30 photons / s
Explanation:
Let's start by finding the energy of a photon of light, let's use the Planck relation
E = h f
the speed of light is related to wavelength and frequency
c = λ f
we substitute
E = h c /λ
E₀ = 6.63 10⁻³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻¹⁹ J
now let's use a direct proportion rule. If the energy of a photon is Eo, how many fornes has an energy E = 1.2 10⁻¹⁷ J in a second
#_photons = 1 photon (E / Eo)
#_photons = 1 1.2 10⁻¹⁷ /3.978 10⁻¹⁹
#_photons = 3.0 10¹
#_photons = 30 photons / s
Answer:
Explanation:
Case 1:
mass = m
initial velocity = vo
final velocity = 0
height = y
Use third equation of motion
v² = u² - 2as
0 = vo² - 2 g y
y = vo² / 2g ... (1)
Case 2:
mass = 2m
initial velocity = 2vo
final velocity = 0
height = y '
Use third equation of motion
v² = u² - 2as
0 = 4vo² - 2 g y'
y ' = 4vo² / 2g
y' = 4 y
Thus, the second rock reaches the 4 times the distance traveled by the first rock.
