Answer: first blank- 26.17° C, second- 1.17 °C, third- 30.86°C, and last fourth- 5.86°C
Explanation:
Physics, got it right on edg. 2020:)
To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,
Here
n = Number of node
T = Tension
= Linear density
L = Length
Replacing the values in the frequency and value of n is one for fundamental overtone
Similarly plug in 2 for n for first overtone and determine the value of frequency
Similarly plug in 3 for n for first overtone and determine the value of frequency
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
a)
Y₀ = initial position of the stone at the time of launch = 0 m
Y = final position of stone = 20.0 meters
a = acceleration = - 9.8 m/s²
v₀ = initial speed of stone at the time of launch = 30.0 m/s
v = final speed = ?
Using the equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
v² = 30² + 2 (- 9.8) (20 - 0)
v = 22.5 m/s
b)
Y₀ = initial position of the stone at the time of launch = 0 m
Y = maximum height gained
a = acceleration = - 9.8 m/s²
v₀ = initial speed of stone at the time of launch = 30.0 m/s
v = final speed = 0 m/s
Using the equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
0² = 30² + 2 (- 9.8) (Y - 0)
Y = 46 m
