Answer:
the size are components relative to the whole.
Explanation:
they are particularly good at showing percentage or proportional data
A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.
Answer:
0.4757 mm
Explanation:
Given that:
Load P = 223,000 N
the length of the height of the aluminium column = 1.22 m
the diameter of the aluminum column = 10.2 cm = 0.102 m
The amount that the column has shrunk ΔL can be determined by using the formula:
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
where;
A = πr²
2r = D
r = D/2
r = 0.102/2
r = 0.051
A = π(0.051)²
A = 0.00817
Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:
[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]
ΔL = 4.757 × 10⁻⁴ m
ΔL = 0.4757 mm
Hence; the amount that the column has shrunk is 0.4757 mm
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop
Answer:
v = 1.7 m/s
Explanation:
By applying conservation of energy principle in this situation, we know that:
Loss in Potential Energy of Car = Gain in Kinetic Energy of Car
mgΔh = (1/2)mv²
2gΔh = v²
v = √(2gΔh)
where,
v = velocity of car at top of the loop = ?
g = 9.8 m/s²
Δh = change in height = 45 cm - Diameter of Loop
Δh = 45 cm - 30 cm = 15 cm = 0.15 m
Therefore,
v = √(2)(9.8 m/s²)(0.15 m)
v = 1.7 m/s
HELP ILL MARK BRAINLIEST PLS!!!!
A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.
Answer:
Yes, work has been done on the mud.
Explanation:
Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.