Answer:
The only incorrect statement is from student B
Explanation:
The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.
Let's examine student claims using these rotation periods
Student A. The time for 4 turns around the sun is
t = 4 88
t = 352 / 58.7 Earth days
In this time I make as many rotations on itself each one with a time to = 58.7 Earth days
#_rotaciones = t / to
#_rotations = 352 / 58.7
#_rotations = 6
therefore this statement is TRUE
student B. the planet rotates 6 times around the Sun
t = 6 88
t = 528 s
The number of rotations on itself is
#_rotaciones = t / to
#_rotations = 528 / 58.7
#_rotations = 9
False, turn 9 times
Student C. 8 turns around the sun
t = 8 88
t = 704 days
the number of turns on itself is
#_rotaciones = t / to
#_rotations = 704 / 58.7
#_rotations = 12
True
The only incorrect statement is from student B
2NO2 means that there is 2 oxygen atoms and one nitrogen with two sets of that. So its the third one
The angle of inclination is calculated using sin
function,
sin θ = 5 m / 20 m = 0.25
θ = 14.4775°
<span>The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25 </span>
F net = 49N
<span>Work is product of net force and distance:
W = F net * d = 49 * 20 </span>
<span>Work = 980 J </span>
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
Answer:
B . energy cannot be created or destroyed
