Complete question is;
If the diameter of the black marble is 3.0 cm, and by using the formula for volume, what is a good approximation of its volume?
Answer:
14 cm³
Explanation:
We will assume that this black marble has the shape of a sphere from online sources.
Now, volume of a sphere is given by;
V = (4/3)πr³
We are given diameter = 3 cm
We know that radius = diameter/2
Thus; radius = 3/2 = 1.5 cm
So, volume = (4/3)π(1.5)³
Volume ≈ 14.14 cm³
A good approximation of its volume = 14 cm³
What type of drawing is shown in the figure?
Answer:
Hi, There! My Answer would Be A. Block diagram Or C. panel Diagram!
Hope this helps! ♚♛♕♔ッ✨♚
A concave lens can only form a virtual image. The correct option among all the options that are given in the question is the third option or option "C". Concave lenses are mostly thinner in the middle compared to its edges. I hope that this answer has come to your help.
The magnitude of the electrical force between q2 and q3 is given as a ratio between the product of their charges and the square of the distance of separation.
<h3>What is the magnitude of electrical forces between two charges?</h3>
The magnitude of the electrical force between two charges refers to the attractive or repulsive forces that exists between two charges separated by a given distance in an electric field.
The magnitude of the electrical force, F between the two charges q2 and q3 is given be my the formula below
Therefore, the magnitude of the electrical force between q2 and q3 is given as a ratio between the product of their charges and the square of the distance of separation.
Learn more about electrical force at: brainly.com/question/17692887
#SPJ4
Answer:
24,000 m
Explanation:
First find the rocket's final position and velocity during the first phase in the y direction.
Given:
v₀ = 75 sin 53° m/s
t = 25 s
a = 25 sin 53° m/s²
Find: Δy and v
Δy = v₀ t + ½ at²
Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²
Δy = 7736.8 m
v = at + v₀
v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)
v = 559.0 m/s
Next, find the final position of the rocket during the second phase (as a projectile).
Given:
v₀ = 559.0 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 15945.5 m
The total displacement is:
7736.8 m + 15945.5 m
23682.2 m
Rounded to two significant figures, the maximum altitude reached is 24,000 m.
