<h2> The potential and kinetic energy of airplane are affected by these factors </h2>
Explanation:
When airplane rises up , it requires potential energy . This potential energy can be taken from the kinetic energy of airplane .
Thus if the speed of wind is larger , it can either oppose the motion of velocity or can favour the velocity of airplane . By which its kinetic energy is effected .
If the weight of airplane is changed , it will effect the potential energy required . Thus heavier plane requires higher potential energy for attaining the same height .
Thus these two factor has important role in the flight of airplane .
The result of the Mexican victory was that fallen defenders
became heroes to the cause of Texan independence.<span> The Battle of
the Alamo took place between February 23 and March 6, 1836 and became the
central episode of the Texas
Revolution . After this thirteen-day battle, the
Mexican troops of General President Antonio
Lopez de Santa Anna began an attack on San Antonio de
Bexar, the current San Antonio in Texas. The Battle of the Alamo fought the
army of Mexico against
a group of Texan rebels, mostly American settlers. More than four thousand
men from Santa Ana stood in front of
the Alamo Fort , the last stronghold of the rebels, which
barely reached 187. The Alamo was not a fortress prepared to withstand a siege.
It is believed that all the rebels of the Alamo died in the siege, but Santa
Anna came to lose up to about 900 men during the days that lasted the fight. However,
the worst result for Santa Ana was precisely the resistance that the Texan
rebels had in the Alamo, which fostered the fighting spirit of the Texans. A
few days later, on March 14, 1836, Texas became independent from Mexico and a
month later, Santa Ana was imprisoned.</span>
Air resistance force
tension force
spring force
frictional force
normal force
gravitational force
applied force
please give me brainly:)
Answer:
62.8 μC
Explanation:
Here is the complete question
The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?
Solution
The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ
So, Q = ∫∫∫ρdV
Q = ∫∫∫ρr²sinθdθdrdΦ
Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ
Q = ∫∫∫0.2r⁴sinθdθdrdΦ
We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π
So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ
Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ
Q = ∫∫0.2r⁴[-cosθ]drdΦ
Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ
Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ
Q = ∫∫0.2r⁴-[- 2]drdΦ
Q = ∫∫0.2r⁴(2)drdΦ
Q = ∫∫0.4r⁴drdΦ
Q = ∫0.4r⁴dr∫dΦ
Q = ∫0.4r⁴dr[Φ]
Q = ∫0.4r⁴dr[2π - 0]
Q = ∫0.4r⁴dr[2π]
Q = ∫0.8πr⁴dr
Q = 0.8π∫r⁴dr
Q = 0.8π[r⁵/5]
Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]
Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]
Q = 0.8π[0.025 m⁵ - 0 m⁵]
Q = 0.8π[0.025 m⁵]
Q = (0.02π mC/m⁵) m⁵
Q = 0.0628 mC
Q = 0.0628 × 10⁻³ C
Q = 62.8 × 10⁻³ × 10⁻³ C
Q = 62.8 × 10⁻⁶ C
Q = 62.8 μC
Answer:
Explanation:
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