Answer:
Their experimental design lacks control group
Explanation:
Based on what I read, the scientists don't have a control group as one of the main groups thus they cannot, in scientific sense, say that the medicine is better or worse. You always need a control group receiving no intervention because then we can compare groups and assess the effectiveness of that intervention (in our case if we had control group vs people who received the medicine, we could see if the people who received the medicine had improved condition etc when compared to participants who did not receive anything)
Answer:
0.37sec
Explanation:
Period of oscillation of a simple pendulum of length L is:
T
=
2
π
×
√
(L
/g)
L=length of string 0.54m
g=acceleration due to gravity
T-period
T = 2 x 3.14 x √[0.54/9.8]
T = 1.47sec
An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.
The ball will first have V(max) at T/4,
=>V(max) = 1.47/4 = 0.37 sec
Answer:
The funda mental frequency of the original tube is 182Hz.
Explanation:
See the attachment for the calculation steps.
In order to calculate the fundamental frequency of the original closed tube we need to find the length of the tube which is equal to the sum of the lengths of the two new tubes.
For closed tubes
f = nv/4L (n = 1, 3, 5,...n)
f = nv/2L (n = 1, 2, 3,...n)
The details of calculation can be found below in the attachment.
Answer:
C. a rolling bowling ball
I just answered this question on my quiz.
Answer:
a. L = μ₀AN²/l b. 1.11 × 10⁻⁷ H
Explanation:
a. The magnetic flux through the solenoid, Ф = NAB where N = number of turns of solenoid, A = cross-sectional area of solenoid and B = magnetic field at center of solenoid = μ₀ni where μ₀ = permeability of free space, n = number of turns per unit length = N/l where l = length of solenoid and i = current in solenoid.
Also, Li = Ф where L = inductance of solenoid.
So, Li = NAB
= NA(μ₀ni)
= NA(μ₀Ni/l)
Li = μ₀AN²i/l
dividing both sides by i, we have
So, L = μ₀AN²/l
b. The self- inductance, L = μ₀AN²/l where
A = πd²/4 where d = diameter of solenoid = 0.150 cm = 1.5 × 10⁻³ m, N = 50 turns, μ₀ = 4π × 10⁻⁷ H/m and l = 5.00 cm = 5 × 10⁻² m
So, L = μ₀AN²/l
L = μ₀πd²N²/4l
L = 4π × 10⁻⁷ H/m × π(1.5 × 10⁻³ m)²(50)²/(4 × 5 × 10⁻² m)
L = 11,103.3 × 10⁻¹¹ H
L = 1.11033 × 10⁻⁷ H
L ≅ 1.11 × 10⁻⁷ H
