Answer:
The frequency of the standing wave in the second case is higher than that in the first case
Explanation:
The frequency and wavelength of a wave are related.
The moment you sliced the bottle, you've reduced the wavelength of the bottle.
When wavelength decreases, frequency increases and vice versa.
So, When frequency
increases in the second case, more wave crests pass a fixed point each second. That means
the wavelength shortens. So, as frequency increases, wavelength
decreases. The opposite is also true—as frequency decreases,
wavelength increases.
Answer:
L= 1 m, ΔL = 0.0074 m
Explanation:
A clock is a simple pendulum with angular velocity
w = √ g / L
Angular velocity is related to frequency and period.
w = 2π f = 2π / T
We replace
2π / T = √ g / L
T = 2π √L / g
We will use the value of g = 9.8 m / s², the initial length of the pendulum, in general it is 1 m (L = 1m)
With this length the average time period is
T = 2π √1 / 9.8
T = 2.0 s
They indicate that the error accumulated in a day is 15 s, let's use a rule of proportions to find the error is a swing
t = 1 day (24h / 1day) (3600s / 1h) = 86400 s
e= Δt = 15 (2/86400) = 3.5 104 s
The time the clock measures is
T ’= To - e
T’= 2.0 -0.00035
T’= 1.99965 s
Let's look for the length of the pendulum to challenge time (t ’)
L’= T’² g / 4π²
L’= 1.99965 2 9.8 / 4π²
L ’= 0.9926 m
Therefore the amount that should adjust the length is
ΔL = L - L’
ΔL = 1.00 - 0.9926
ΔL = 0.0074 m
Could you supply u with some answer choices please?
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Answer
given,
F₁ is horizontal = 40 N
F₂ is normal = 20 N
F₃ is parallel = 30 N
work done by
W₂ = 0 as force is acting perpendicular to the direction of motion.
as the motion moved to 0.8 cm
W₃ = F₃ x d
W₃ = 30 x 0.8
W₃ = 24 J
W₁ = F₁ x d
W₁ = F₁ cos ∅ x d
W₁ = 40 cos 30⁰ x 0.8
W₁ = 27.21 J
