Answer:
Density relates a mass to its volume.
Density varies with temperature
Density determines if a substance floats or sinks.
Density may have units of grams per milliliter (g/mL)
Explanation:
Density 
is a characteristic property of a substance or material and is defined as the relationship between the mass 
of a body or substance and the volume 
it occupies:
This means the density is inversely proportional to the volume.
On the other hand, density is a scalar quantity and according to the International System of Units its unit is 
, although it can be also expressed in 
.
It should be noted that the density of a body is related to its buoyancy, a substance or body will float on another fluid if its density is lower. In addition, if the pressure of the substance remains constant, as the temperature increases, the density decreases; this means density varies with the temperature as well.
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:
At the instant shown in the diagram, the car's centripetal acceleration is directed is discussed below in detail.
Explanation:
The direction of the centripetal acceleration is in a circular movement is forever towards the middle of the roundabout pathway. In the picture displayed, the East direction is approaching the center. So, the course of the car's centripetal acceleration is (H) toward the east.
<h3><u>Question</u><u>:</u></h3>
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?
<h3><u>Statement:</u></h3>
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.
<h3><u>Solution</u><u>:</u></h3>
- Initial velocity (u) = 70 m/s
- Acceleration (a) = -14 m/s^2
- Time (t) = 3 s
- Let the velocity of the car after 3 s be v m/s
- By using the formula,
v = u + at, we have
- So, the velocity of the car after 3 s is 28 m/s.
<h3><u>Answer:</u></h3>
The car's speed after 3 s is 28 m/s.
Hope it helps
