Answer:
A. a parked bus
Explanation:
Because a parked bus probably has the most mass out of these 4 and as we know, mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.
Answer:
2m₁m₃g / (m₁ + m₂ + m₃)
Explanation:
I assume the figure is the one included in my answer.
Draw a free body diagram for each mass.
m₁ has a force T₁ up and m₁g down.
m₂ has a force T₁ up, T₂ down, and m₂g down.
m₃ has a force T₂ up and m₃g down.
Assume that m₁ accelerates up and m₂ and m₃ accelerate down.
Sum of the forces on m₁:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of the forces on m₂:
∑F = ma
T₁ − T₂ − m₂g = m₂(-a)
T₁ − T₂ − m₂g = -m₂a
(m₁g + m₁a) − T₂ − m₂g = -m₂a
m₁g + m₁a + m₂a − m₂g = T₂
(m₁ − m₂)g + (m₁ + m₂)a = T₂
Sum of the forces on m₃:
∑F = ma
T₂ − m₃g = m₃(-a)
T₂ − m₃g = -m₃a
a = g − (T₂ / m₃)
Substitute:
(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂
(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂
(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂
m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂
2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂
T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)
The distance an object falls, from rest, in gravity is
D = (1/2) (G) (T²)
'T' is the number seconds it falls.
In this problem,
0.92 meter = (1/2) (9.8) (T²)
Divide each side by 4.9 : 0.92 / 4.9 = T²
Take the square root
of each side: √(0.92/4.9) = T
0.433 sec = T
The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor. BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.
The pencil is in the air for 0.433 second.
In that time, it travels
(0.433s) x (1.4 m/s) = 0.606 meter
from the edge of the table.
Answer:
A. 0.289g/mL
Explanation:
Using the equation for density which is d = m/v or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.
The tension in the upper rope is determined as 50.53 N.
<h3>Tension in the upper rope</h3>
The tension in the upper rope is calculated as follows;
T(u) = T(d)+ mg
where;
- T(u) is tension in upper rope
- T(d) is tension in lower rope
T(u) = 12.8 N + 3.85(9.8)
T(u) = 50.53 N
Thus, the tension in the upper rope is determined as 50.53 N.
Learn more about tension here: brainly.com/question/918617
#SPJ1
