Answer:
T = 4.42 10⁴ N
Explanation:
this is a problem of standing waves, let's start with the open tube, to calculate the wavelength
λ = 4L / n n = 1, 3, 5, ... (2n-1)
How the third resonance is excited
m = 3
L = 192 cm = 1.92 m
λ = 4 1.92 / 3
λ = 2.56 m
As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio
v = λ f
f = v / λ
f = 343 / 2.56
f = 133.98 Hz
Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m
The expression for standing waves on a string is
λ = 2L / n
λ = 2 0.37 / 2
λ = 0.37 m
The speed of the wave is
v = λ f
As we have some resonance processes between the string and the tube the frequency is the same
v = 0.37 133.98
v = 49.57 m / s
Let's use the relationship of the speed of the wave with the properties of the string
v = √ T /μ
T = v² μ
T = 49.57² 18
T = 4.42 10⁴ N
Answer:
The child will take 5.952 seconds to travel from the top of the hill to the bottom.
Explanation:
Given that the child accelerates uniformly and that both initial (
) and final speeds (
), measured in meters per second, and acceleration (
), measured in meters per square second, are known, we proceed to use the following kinematic equation to determine the time taken to travel from the top of the hill to the bottom (
), measured in seconds, is:
(1)
If we know that 
, 
and 
, then the time taken is:
The child will take 5.952 seconds to travel from the top of the hill to the bottom.
Answer:
If a negatively charged balloon is brought near one end of the rod but not in direct contact, then <u>the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon</u>, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.
A peak = A Rms x Sq root 2
Therefore 3.6 x sq root of 2
A peak = 5.09
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.
