A 1.0-kilogram ball is dropped from the roof of a building 40. meters tall. What is the approximate time of fall? [Neglect air r
2 answers:
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Answer:
(a) p = 3.4 kg-m/s (b) 37.78 N.
Explanation:
Mass of a basketball, m = 0.4 kg
Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)
It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)
(a) Change in momentum = final momentum - initial momentum
p = m(v-u)
p = 0.4 (2.8-(-5.7))
p = 3.4 kg-m/s
(b) Impulse = change in momentum
Ft = 3.4
We have, t = 0.09 s
Hence, this is the required solution.
<h2>Given :</h2>
- total charge = 9.0 mC = 0.009 C
Each electron has a charge of :
For producing 1 Cuolomb charge we need :
Now, for producing 0.009 C of charge, the number of electrons required is :
_____________________________
So, Number of electrons passing through the cross section in 3.6 seconds is :
Number of electrons passing through it in 1 Second is :
Now, in 10 seconds the number of electrons passing through it is :
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Answer:
a) f = 4.76 10¹⁴ Hz, b) d = 2.73 10⁻⁴ m, c) θ = 6.923 10⁻³ rad
Explanation:
a) In this problem the frequency of light is asked, let's use the relationship between the speed of the wave, its wavelength and its frequency
c = λ f
f = c /λ
f =
f = 4.76 10¹⁴ Hz
b) slit separation (d)
the expression for the constructive interference of the double-slit experiment is
d sin θ = m λ
let's use trigonometry
tan θ = y / L
tan θ =
in general the angles are small, so we can approximate
tan θ = sin θ
tan θ = y/L
we substitute
d y / L = m λ
d = m L λ / y
we calculate
d = 3 1.3 630 10⁻⁹ /0.90 10⁻²
d = 2.73 10⁻⁴ m
c) the angle
tan θ = y / L
θ = tan⁻¹ y / L
θ = tan⁻¹ 0.9 10⁻² / 1.3
θ = tan⁻¹ 6,923 10⁻³
let's find the angle in radians
θ = 6.923 10⁻³ rad