atom ................................................... sorry for the periods
Answer:
a. Both wires have the same resistivity
Explanation:
For the resistance of a wire , following formula holds good .
R = ρ l / S , R is resistance , l is length , S is cross sectional area and ρ is resistivity of the material that the wire is made of. Resistance is dependent on length and cross sectional area but resistivity does not depend upon length or cross sectional area . It only depends upon the type of material.
If we replace copper wire with aluminium wire , then resistivity will change .
Hence , since the wire remains made of copper , resistivity will not change.
Answer:
Shiny metals such as copper, silver, and gold are often used for decorative arts, jewelry, and coins.
Strong metals such as iron and metal alloys such as stainless steel are used to build structures, ships, and vehicles including cars, trains, and trucks.
Some metals have specific qualities that dictate their use. For example, copper is a good choice for wiring because it is particularly good at conducting electricity. Tungsten is used for the filaments of light bulbs because it glows white-hot without melting.
Nonmetals are plentiful and useful. These are among the most commonly used:
Oxygen, a gas, is absolutely essential to human life. Not only do we breathe it and use it for medical purposes, but we also use it as an important element in combustion.
Sulfur is valued for its medical properties and as an important ingredient in many chemical solutions. Sulfuric acid is an important tool for industry, used in batteries and manufacturing.
Chlorine is a powerful disinfectant. It is used to purify water for drinking and fill swimming pools.
Explanation:
Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
With constant angular acceleration 
, the disk achieves an angular velocity 
at time 
according to
and angular displacement 
according to
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
b. Under constant acceleration, the average angular velocity is equivalent to
where 
and 
are the final and initial angular velocities, respectively. Then
c. After 1.00 s, the disk has instantaneous angular velocity
d. During the next 1.00 s, the disk will start moving with the angular velocity 
equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle according to
which would be equal to
