Answer:
Kinetic Energy
Explanation:
Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force. ... Kinetic energy can be transferred between objects and transformed into other kinds of energy. For example, a flying squirrel might collide with a stationary chipmunk.
Answer:
sex korogi tum mere to mere
Respuesta: verifique amablemente la explicación
Explicación:
Dado lo siguiente:
Longitud (L) del cable = 120 m
Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m
Fuerza (F) = 380 N
Esfuerzo longitudinal = Fuerza / Área
Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4
Área = (3.142 * 4.84 * 10 ^ -6)
Área = 0.00000380132 m²
Estrés = Fuerza / Área
Estrés = 380 / 0.00000380132
Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2
Deformación longitudinal: extensión / longitud
Extensión = 0.10 m
Longitud = 120 m
Deformación longitudinal = 0,1 m / 120 m
Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4
Part a:
= 56
= 60
= 63
The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The is the overall medium, is the medium of the first half, and is the medium of the second half.
-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.
[] See attached
Part b:
The range is 7.
The interquartile range is the range of numbers between and . In other words, it is 50% of the data, directly in the middle.
This becomes 63 - 56 = 7
Part c:
79 is an outlier.
It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.
-> 63 + (7 + 
) ≤ 79
-> 63 + 10.5 ≤ 79
-> 73.5 ≤ 79
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.
- Heather
Answer:
B. d(low)=4d(high)
Explanation:
Frequency of a string can be written as;
f = v/2L
Where;
v = sound velocity
L = string length
Frequency can be further expanded to;
f = v/2L = (1/2L)√(T/u) ......1
Where;
m= mass,
u = linear density of string,
T = tension
p = density of string material
A = cross sectional area of string
d = string diameter
u = m/L .......2
m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)
f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))
f = (1/2L)√(T/((p(πd^2)L/4)/L))
f = (1/2L)√(4T/pπd^2)
f = (1/L)(1/d)√(4T/pπ)
Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.
f ~ 1/d
So, if
4f(low) = f(high)
Then,
d(low) = 4d(high)
