The figures below show systems containing a block initially held at rest on a frictionless surface. In each system, the block is
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Answer:
F_scale = 20.18 N
Explanation:
The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.
* Let's find the amount of liquid in the container for a time of t = 2.93 s
Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there
V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l
V_{water} = 1.09 10⁻³ m³
the amount of water is
ρ = m / V
m = ρ V
m = 1000 1.09 10⁻³
m = 1.09 kg
so the weight of the liquid in the container for this time is
W = mg
W = 1.09 9.8
W = 10.68 N
* Let's look for the force of the falling jet
Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h
P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2 ρ g v₂² + ρ g y₂
In this case, the water falls freely, so the external pressure is atmospheric.
P₂ = P_{atm}
since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0
let's use kinematics to find the speed of a drop when it reaches the container y = 0
v² = v₀² - 2 g (y-y₀)
v =
let's calculate
v = √(2 9.8 40.5)
v = 28.17 m / s
this is the speed in the container v₁ = 28.17 m / s
the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0
we substitute in Bernoulli's equation
P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂
P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂
P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²
P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²
P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵
P₁ = 1.28 10⁵ Pa
The definition of Pressure is
P = F / A
F = P A
We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving
m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³
the volume is V = A l
if the length of l = 1 m
A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²
the force of this jet is
F = P A
F = 1.28 10⁵ 7.46 10⁻⁵
F = 9.5 N
with these data let's use the equilibrium equation
F_ scale -W - F = 0
F_scale = W + F
F_scale = 10.682 + 9.5
F_scale = 20.18 N
Car is moving on the glassy slope with constant speed
Now we know that
so acceleration is rate of change in velocity
as we know that velocity is constant here so acceleration is zero
so here
now as we know by Newton's II law
since a = 0
so net force will be ZERO on it during this motion
Answer:
t=15.68 s
Explanation:
Given that
m = 1520 kg
P =75 KW
We know that
Power ,P = F .v
F=force
v=velocity
v= 100 km/h
v=27.77 m/s
75 x 1000 = F x 27.77
F= 2700.75 N
F= m a
m=mass
a=acceleration
2700.75 = 1520 x a
a=1.77 m/s²
time t given as
v= u + a t
27.77 = 0 + 1.77 x t
t=15.68 s