Answer:
2.72 cycles
Explanation:
First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is
The period of the pendulum instead is given by:
Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:
Answer:
Explanation:
During a car collision momentum of vehicle ceases within a fraction of seconds so Force due to the impulse is huge.
Impulse is defined as the product of average force and time. If we can increase the period of collision for the same impulse then the average force imparted will be less.
If we can increase the time period then damage due to collision will be less.
Answer:
0.82 mm
Explanation:
The formula for calculation an 
bright fringe from the central maxima is given as:
so for the distance of the second-order fringe when wavelength 
= 745-nm can be calculated as:
where;
n = 2
= 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
= 0.00276 m
= 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength 
= 660-nm is as follows:
= 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
= 10 ⁻³ (2.76 - 1.94)
= 10 ⁻³ (0.82)
= 0.82 × 10 ⁻³ m
= 0.82 × 10 ⁻³ m 
= 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
Answer:
The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.
Explanation:
The photoelectric effect occurs when light is shined on metals such as zinc beyond a certain frequency (the threshold frequency), which causes electrons to escape from the zinc. The electrons which are fleeing are called photo electrons.
Therefore photo electric effect is
The photoelectric effect occurs only for frequencies above the cutoff frequency, regardless of the intensity.
Answer:
we got time and velocity over time.
so the distance is again the area underneath the graph
for a triangle with known base and height it's
4*10 / 2
distance traveled is 20
deceleration occurs when velocity decreases. that happens from t=2 till t=4
in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time
a (from t=2 to t=4) = -5v/t
