Answer:

Explanation:
Note that acceleration is the rate change of velocity i.e
.
Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have


Answer:
La frecuencia cardíaca objetivo durante las actividades de intensidad moderada es aproximadamente del 50 al 70% de la frecuencia cardíaca máxima, mientras que durante la actividad física intensa es de, aproximadamente, entre el 70 y el 85% del valor máximo.
Answer:
x = 727.5 km
Explanation:
With the conditions given using trigonometry, we can find the tangent
tan θ = CO / CA
With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun
D =150 10⁶ km (1000m / 1 km)
D = 150 10⁹ m.
We must take the given angle to radians.
1º = 3600 arc s
π rad = 180º
θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =
θ = 4.85 10⁻⁶ rad
That angle is extremely small, so we can approximate the tangent to the angle
θ = x / D
x = θ D
x = 4.85 10-6 150 109
x = 727.5 103 m
x = 727.5 km
Answer:
Explanation:
From the given information:
We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then
The volume charge distribution relates to the radial direction at r = R
∴



To find the constant k, we examine the total charge Q which is:


∴



Thus;




Hence, from equation (1), if k = 


To verify the units:

↓ ↓ ↓
c/m³ c/m³ × 1/m
Thus, the units are verified.
The integrated charge Q



since 

Answer:
Explanation:
The formula for this is
where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.
Filling in:
and moving things around to solve for r:
Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m