Answer:
(a) FN = m (g -
)
(b) vmin = 17.146 m/s
Explanation:
The radius of the arc is
r = 30m
The normal force acting on the car form the highest point is
FN = m (g -
)
If the normal force become 0 we have
m (g -
) = 0
or
g -
= 0
This way, when FN = 0, then v = vmin, so
g -
= 0
vmin =
= ![\sqrt[.]{9.8 m/s^{2} * 30m } = 17.146 m/s](https://tex.z-dn.net/?f=%5Csqrt%5B.%5D%7B9.8%20m%2Fs%5E%7B2%7D%20%2A%2030m%20%7D%20%3D%2017.146%20m%2Fs)
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 final potential energy of this system is 3U0/10
Explanation:
We are given
charge at left end and another test charge at point p
Potential energy is given by =
where k is electrostatics constant = 
Q1 = first charge , Q2= test charge
R= distance between charges
potential at point p
U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1
now the test charge moves to point R
using Pytahgoreou theorem
R(distance) =
= 10
New Potential energy
U1 = kq1*q2 / 10
substituting kq1q2 = 3U0 from 1
U1 = 3U0/10
So this is the final potential energy of this system.
Answer & Explanation:
Relative age is the age of a rock layer (or the fossils it contains) compared to other layers. It can be determined by looking at the position of rock layers. Absolute age is the numeric age of a layer of rocks or fossils.
I changed my undershorts. The elastic on the old ones I put on that day was deteriorated, and it completely failed when I dripped lab coffee on it, causing falldown.