The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive towa
1 answer:
You might be interested in
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
Answer:
The volume of the submerged part of her body is
Explanation:
Let's define the buoyant force acting on a submerged object.
In a submerged object acts a buoyant force which can be calculated as :
ρ.V.g
Where ''B'' is the buoyant force
Where ''ρ'' is the density of the fluid
Where ''V'' is the submerged volume of the object
Where ''g'' is the acceleration due to gravity
Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.
We can write :
(I)
Where ''W'' is weight
⇒ If we consider ρ = (water density) and
and replacing this values in the equation (I) ⇒
ρ.V.g = 610N
(II)
The force unit ''N'' (Newton) is defined as
Using this in the equation (II) :
We find that the volume of the submerged part of her body is