4. Un móvil lleva una velocidad de 75 m/s, en dicho momento aplica los frenos durante tiempo de 15 segundos hasta detenerse. ¿cu
1 answer:
Answer:
a = 5 m / s² and x = 562.5 m
Explanation:
Let's apply the kinematics relations
v = v₀ + at
v² = v₀² + 2a x
In this case, as the car stops, its final speed is zero (v = 0) and the acceleration must be opposite to the speed, that is, negative
0 = v₀ - a t
a = v₀ / t
let's calculate
a = 75/15
a = 5 m / s²
let's look for the distance
0 = v₀² - 2a x
x = v₀² / 2a
let's calculate
x =
x = 562.5 m
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Answer:
The depth to which the hydrometer sinks is approximately 24.07 cm
Explanation:
The given parameters are;
The diameter of the hydrometer tube, d = 2.3 cm
The mass of the content of the tube, m = 80 g
The density of the liquid in which the tube floats, ρ = 800 kg/m³
By Archimedes' principle, the up thrust (buoyancy) force acting on the hydrometer = The weight of the displaced liquid
When the hydrometer floats, the up-thrust is equal to the weight of the hydrometer which by Archimedes' principle, is equal to the weight of the volume of the liquid displaced by the hydrometer
Therefore;
The weight of the liquid displaced = The weight of the hydrometer, W = m·g
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ W = 80 g × g
The volume of the liquid that has a mass of 80 g (0.08 kg), V = m/ρ
V = 0.08 kg/(800 kg/m³) = 0.0001 m³ = 0.0001 m³ × 1 × 10⁶ cm³/m³ = 100 cm³
The volume of the liquid displaced = 100 cm³ = The volume of the hydrometer submerged,
= A × h
Where;
A = The cross-sectional area of the tube = π·d²/4
h = The depth to which the hydrometer sinks
h = /A
∴ h = 100 cm³/( π × 2.3²/4 cm²) ≈ 24.07 cm
The depth to which the tube sinks, h ≈ 24.07 cm.
Recall this gas law:
=
P₁ and P₂ are the initial and final pressures.
V₁ and V₂ are the initial and final volumes.
T₁ and T₂ are the initial and final temperatures.
Given values:
P₁ = 475kPa
V₁ = 4m³, V₂ = 6.5m³
T₁ = 290K, T₂ = 277K
Substitute the terms in the equation with the given values and solve for Pf:
The formula for velocity vf = vi + at
First list your given information
2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)
Since you want the a for acceleration get a by itself
a = (vf-vi)/t
So a= (6-2)/2
a= 4/2
a=2
Now units
the units for acceleration are m/s
2m/s
First list your given information
2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)
Since you want the a for acceleration get a by itself
a = (vf-vi)/t
So a= (6-2)/2
a= 4/2
a=2
Now units
the units for acceleration are m/s
2m/s