Answer:
Explanation:
F = ma
<u>Assuming</u> the 20° is angle θ measured to the horizontal
mgsinθ - μmgcosθ = ma
g(sinθ - μcosθ) = a
at constant velocity, a = 0
g(sinθ - μcosθ) = 0
sinθ - μcosθ = 0
sinθ = μcosθ
μ = sinθ/cosθ
μ = tanθ
μ = tan20
μ = 0.3639702342...
μ = 0.36
Answer:
0.29 m/s due west.
Explanation:
According to newton's second law,
Net force acting on an object = mass×acceleration
From the question,
F+F₁+F₂ = ma................ Equation 1
Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.
Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg
substitute into equation 1
4080-680-1160 = 7660(a)
2240 = 7660a
Therefore,
a = 2440/7660
a = 0.29 m/s due west.
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 =
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = <u> mass </u>
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg
Answer:
superscript
Explanation:
When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.
An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.