Answer:
v₀ = 13.9 10³ m / s
Explanation:
Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.
F = m a
G m M / x² = m dv / dt = m dv/dx dx/dt
G M / x² = dv/dx v
GM dx / x² = v dv
We integrate
v² / 2 = GM (-1 / x)
We evaluate between the lower limits where x = Re = 6.37 10⁶m and the velocity v = vo and the upper limit x = 2.50 10⁸m with a velocity of v = 8.50 10³ m/s
½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))
72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)
72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3 10⁻⁸)
72.25 10⁶ - v₀² = -1.213 10⁸
v₀² = 72.25 10⁶ + 1,213 10⁸
v₀² = 193.6 10⁶
v₀ = 13.9 10³ m / s
Answer:
acceleration = 0.2625 m/s²
Explanation:
acceleration = ( final velocity - initial velocity ) / time
Here the final velocity is 10.6 m/s and initial velocity is 6.4 m/s and time is 16 s.
using the equation:
acceleration = ( 10.6 - 6.4 ) / 16
= 0.2625 m/s²
Explanation:
Fgravity = G*(mass1*mass2)/D².
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between the objects.
so, if you triple one of the masses, what does that do to our equation ?
Fgravitynew = G*(3*mass1*mass2)/D²
due to the commutative property of multiplication
Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity
so, the right answer is 3×12 = 36 units.
Answer:
Explanation:
Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.
P = h d g
Where, h is the depth, d be the density and g be the acceleration due to gravity.
If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.
Answer: Approximately 65% from what i have learnt.
