Explanation:
It is given that,
Mass of the brick, m = 1.15 kg
Radius of the circle, r = 1.44 m
The cable will break if the tension exceeds 43.0 N
Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,
v = 6.30 m/s
So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.
Use the kinematic equation: Vf=Vi+at
Then plug;
Vi=14 m/s
a=5 m/s²
t=20 s. Therefore;
Vf=14+(5*20)
Vf=114 m/s.
Answer:
(1) 0.333 Hz
(2) 4 sec
(3) 2 sec, 0.5 Hz
Explanation:
(1) We have given time period of pendulum is 3 sec
So T = 3 sec
Frequency will be equal to 
(2) Frequency of the pendulum is given f = 0.25 Hz
Time period is equal to 
(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds
So time taken to complete 1 back and forth swings = 
So time period T = 2 sec
Frequency will be equal to 
Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3 + CaCO3⬇. NaNO3 is solution so CaCO3 is the precipitate formed.
Answer:
The lowest possible frequency of sound for which this is possible is 1307.69 Hz
Explanation:
From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.
First, we will determine his distance from the second speaker using the Pythagorean theorem
l₂ = √(2.00²+5.00²)
l₂ = √4+25
l₂ = √29
l₂ = 5.39 m
Hence, the path difference is
ΔL = l₂ - l₁
ΔL = 5.39 m - 5.00 m
ΔL = 0.39 m
From the formula for destructive interference
ΔL = (n+1/2)λ
where n is any integer and λ is the wavelength
n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.
Then,
0.39 = (1+ 1/2)λ
0.39 = (3/2)λ
0.39 = 1.5λ
∴ λ = 0.39/1.5
λ = 0.26 m
From
v = fλ
f = v/λ
f = 340 / 0.26
f = 1307.69 Hz
Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.
