There’s no picture so how r we supposed to answer it
Missing figure and missing details can be found here:
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Solution:
(a) The work done by the spring is given by

where k is the elastic constant of the spring and

is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:

where the negative sign is given by the fact that

points in the opposite direction of the displacement of the cart, and where

therefore, the work done by the weight is
In naming covalent compound (binary) based in IUPAC naming, we have 4 rules to be followed:
1. The first element of the formula will use the normal name of the given element. for example: CO2 ( Carbon Dioxide), Carbon is the element name of the first element of the formula.
2. The second element is named as if they are treated like an anion but put in mind that these are no ions in a covalent compound but we put -ide on the second element as if it is an anion.
3. Prefixes are used to indicate the number of atom of the elements in the compound. for example: mono- 1 atom, di- 2atoms, tri- 3 atoms and etc
4. Prefix "mono"is never used in naming the first element. For example: Carbon dioxide, there should be no monocarbon dioxide.
Answer:
R₂ / R₁ = D / L
Explanation:
The resistance of a metal is
R = ρ L / A
Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section
We apply this formal to both configurations
Small face measurements (W W)
The length is
L = W
Area
A = W W = W²
R₁ = ρ W / W² = ρ / W
Large face measurements (D L)
Length L = D= 2W
Area A = W L
R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L
The relationship is
R₂ / R₁ = 2W²/L
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .