Answer:
55.66 m
Explanation:
While falling by 50 m , initial velocity u = 0
final velocity = v , height h = 50 , acceleration g = 9.8
v² = u² + 2gh
= 0 + 2 x 9.8 x 50
v = 31.3 m /s
After that deceleration comes into effect
In this case final velocity v = 17 m/s
initial velocity u = 31.3 m/s
acceleration a = - 61 m/s²
distance traveled h = ?
v² = u² + 2gh
(17)² = (31.3)² - 2x 61xh
h = 690.69 / 2 x 61
= 5.66 m
Total height during which he was in air
= 50 + 5.66
= 55.66 m
Answer:
13.51 nm
Explanation:
To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians
y/L=tan θ ≈ θ
and ∆θ ≈∆y/L
Where ∆y= wavelength distance= 2.92 mm =0.00292m
L=screen distance= 2.40 m
=0.00292m/2.40m
=0.001217 rad
The grating spacing is d = (90000 lines/m)^−1
=1.11 × 10−5 m.
the small-angle
approx. Using difraction formula with m = 1 gives:
mλ = d sin θ ≈ dθ →
∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad
=0.000000001351m
= 13.51 nm
Answer:

Explanation:
<u>Motion with Constant Acceleration</u>
A body moves with constant acceleration when the speed changes uniformly in time. The equation used to find the final speed vf is

Where vo is the initial speed, a is the acceleration, and t is the time.
The cyclist has an initial speed of vo=10 miles/hour and ends up at vf=20 miles/hour in t=5 seconds.
Both speeds are given in miles/hour and we must convert it to m/s:
1 mile/hour = 0.44704 m/s
10 mile/hour = 4.47 m/s
20 mile/hour = 8.94 m/s
The acceleration is calculated by solving for a:



Answer:
Amplitude—distance between the resting position and the maximum displacement of the wave
Frequency—number of waves passing by a specific point per second
Period—time it takes for one wave cycle to complete
wavelength λ - the distance between adjacent identical parts of a wave, parallel to the direction of propagation.
Tension - described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar one-dimensional continuous object, or by each end of a rod, truss member, or similar three-dimensional object