Sulfur trioxide
Carbon monoxide
Explanation:
Molecule A is SO₃
Molecule B is CO
SO₃ is sulfur trioxide formed usually by the reaction of SO₂ with oxygen. The bond type between the molecule is a covalent bond because two non-metals with low electronegativity differences are involved.
Here three oxygen atoms are attached to one sulfur atom.
CO is carbon monoxide. A poisonous gas with a wide range of industrial use. A covalent bond exists between the two atoms.
Learn more:
Molecules brainly.com/question/7213980
#learnwithBrainly
<span>Density = mass/volume
10g/2mL=5g/mL
Atomic mass= protons + neutrons
Atomic number=protons
In a neutral atom, electrons = protons
15 amu-7 atomic number = 8 neutrons, 7 protons and 7 electrons</span>
Answer:
The statement which explains how the total time spent in the air is affected as the projectile's angle of launch increases from 25 degrees to 50 degrees is;
C. Increasing the angle from 25° to 50° will increase the total time spent in the air
Explanation:
The equation that can be used to find the total time, T, spent in the air of a projectile is given as follows;

Where;
T = The time of flight of the projectile = The time spent in the air
u = The initial velocity of the projectile
θ = The angle of launch of the projectile
g = The acceleration due to gravity ≈ 9.81 m/s²
Given that sin(50°) > sin(25°), when the angle of launch, θ, is increased from 25 degrees to 50 degrees, we have;
Let T₁ represent the time spent in the air when the angle of launch is 25°, and let T₂ represent the time spent in the air when the angle of launch is 50°, we have;


sin(50°) > sin(25°), therefore, we have;

Therefore;
T₂ > T₁
Therefore, increasing the angle at which the projectile is launched from 25° to 50° will increase the total time spent in the air.
Answer: The vertical displacement will be 127.5 meters
Step by step:
Since the question asks for the vertical displacement only, it is sufficient to limit our calculation to the vertical components of the problem.
We choose to measure the displacement on a vertical axis d with an initial coordinate 0 coinciding with the launch of the ball. There is an initial horizontal velocity v_0 of 50 m/s and the gravitational deceleration g acting in the direction opposite to the initial velocity. The formula for displacement of a body subject to an deceleration with an initial velocity (at time t=5 seconds) is:
After 5 seconds of the launch, the ball will be at the point 127.5 meters from the origin on the vertical axis (vertical displacement)
Metals and properties of metals.