Answer:
1,85 m / s²
Explanation:
De la pregunta anterior, se obtuvieron los siguientes datos:
Velocidad inicial (u) = 40 km / h
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:
1 km / h = 0,2778 m / s
Por lo tanto,
40 km / h = 40 km / h × 0,2778 m / s / 1 km / h
40 km / h = 11,11 m / s
Por tanto, 40 km / h equivalen a 11,11 m / s.
Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:
Velocidad inicial (u) = 11,11 m / s
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
a = (v - u) / (t₂ - t₁)
a = (0 - 11,11) / (6 - 0)
a = - 11,11 / 6
a = –1,85 m / s²
Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²
Answer:
a)-1.014x
J
b)3.296 x
J
Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x
x 47 x 110/ 3.4 => -1.014x
J
b) at d= 0.8m (3.4-2.6) and
=-1.014x
J
The sum of potential and kinetic energies must be conserved as the energy is conserved.
+
=
+ 
As sphere starts from rest and sphere A is fixed at its place, therefore
is zero
=
+ 
The final potential energy is
= - GMaMb/d
Solving for '
'
=
+ GMaMb/d => -1.014x
+ 6.67 x
x 47 x 110/ 0.8
= 3.296 x
J
Answer:
a = 0.55 m / s²
Explanation:
The centripetal acceleration is given by the relation
a = v² / r
angular and linear velocities are related
v = w r
we substitute
a = w² r
In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system
w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s
let's calculate
a = 0.105² 50.0
a = 0.55 m / s²
Answer:
26.9 Pa
Explanation:
We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:
(1)
where
is the cross-sectional area of the 1st section of the pipe
is the cross-sectional area of the 2nd section of the pipe
is the velocity of the 1st section of the pipe
is the velocity of the 2nd section of the pipe
In this problem we have:
is the velocity of blood in the 1st section
The diameter of the 2nd section is 74% of that of the 1st section, so

The cross-sectional area is proportional to the square of the diameter, so:

And solving eq.(1) for v2, we find the final velocity:

Now we can use Bernoulli's equation to find the pressure drop:

where
is the blood density
are the initial and final pressure
So the pressure drop is:

Answer:
Explanation:
A. The earth about its axis:
The earth makes one revolution in 24 hours. to know the number of revolutions per second it makes, we need to convert hours to seconds and the revolution to rad.

B. The minute hand of the clock makes one revolution in 60 minutes
To convert this to rad per second, we have

C. The hour hand of a clock completes one revolution in 12 hours

D. an egg beater turning at 300 rpm.
