a) An inflated balloon was pressed against a wall after it has been rubbed with a piece of synthetic cloth. It was found that the balloon sticks to the wall. <u>This is because a positive and negative electric charge is produced, therefore the balloon sticks to the wall.</u>
b) When an object is thrown up, it comes back to ground <u>because of gravitational attraction force of earth</u>.
c) Mountaineers suffer nose bleeding at higher altitudes <u>because the oxygen level decreases with increase in altitude, which the body cannot adjust.</u>
d) Foundations of high rise buildings are kept wide <u>because more is the area of contact, less is the pressure efforts. So, foundations are wide so as to decrease the possibility of the building from falling down.</u>
e) Deep sea divers or high altitude fliers wear special suits <u>so as prevent their body from being crushed by the water pressure. Since water pressure is maximum at deep seas and oceans, therefore, more is the risk of being injured.</u>
f) Walls of a dam are thickened near the base <u>so that the dam can handle the kinetic energy pressure and prevent itself from breaking down, which if not, can lead to flooding</u>.
HOPE IT HELPS...
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
The AU ... Astronomical Unit ... used to be defined as the average distance between the Sun and Earth during the year.
Now it's defined as 149,597,870,700 meters exactly.
Answer:
The initial velocity of the car is V1= 5.58 m/s = 20.12 km/h
Explanation:
m1= 1689 kg
v1= ?
m2= 2000kg
v2= 17 km/h = 4.72 m/s
by the conservation of quantity of movement :
m1*v1 = m2*v2
v1= m2*v2 / m1
v1= 5.58 m/s = 20.12 km/h
