Answer:
V= 6.974 m/s
Explanation:
Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N
Force of kinetic friction acting parallel and up roof = 18.0 N
Fnet force acting on tool box acting parallel and down roof
Fnet= 55.4 - 18.0
Fnet=37.4 N
acceleration of tool box down roof
a = 37.4(9.81)/88.0
a= 4.169 m/s²
d = 4.90 m
t = √2d/a
t= √2(4.90)/4.169
t= 1.662 s
V = at
V= 4.169(1.662)
V= 6.974 m/s
Answer:
A, the amount of charge stored per volt
Explanation:
Capacitance is defined below:
C = Q/V
Therefore capacitance is charge per volt which gets the unit farad.
Answer:
Part(a): The frequency is 
.
Part(b): The speed of the wave is 
.
Explanation:
Given:
The distance between the crests of the wave, 
.
The time required for the wave to laps against the pier, 
The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is 
.
Also, the time required for the wave for each laps is the time period of oscillation and it is given by 
.
Part(a):
The relation between the frequency and time period is given by
Substituting the value of 
in equation (1), we have
Part(b):
The relation between the velocity of a wave to its frequency is given by
Substituting the value of 
and 
in equation (2), we have
Answer:
P₁- P₂ = 91.1 10³ Pa
Explanation:
For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)
In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference
For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m
P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)
P₁-P₂ = 22.5 10³ + 68.6 10³
P₁- P₂ = 91.1 10³ Pa
