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: 0m/s²
Explanation:
Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane
According newton's law of motion
Summation of forces along the plane = mass × acceleration
Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane
Ff = nR where
n is coefficient of friction = tan(theta)
R is normal reaction = Wcos(theta)
Fm = Wsin(theta)
Substituting in the formula of newton's first law we have;
Fm-Ff = ma
Wsin(theta) - nR = ma
Wsin(theta) - n(Wcos(theta)) = ma... 1
Given
W = 562N, theta = 30°, n = tan30°, m = 56.2kg
Substituting in eqn 1,
562sin30° - tan30°(562cos30°) = 56.2a
281 - 281 = 56.2a
0 = 56.2a
a = 0m/s²
This shows that the trunk is not accelerating
The spine is the one that keeps a huminss body mantaned
Answer:
t should be 3.57 second
Explanation:
Formula used is v = u+at
In which v is final velocity, u is initial velocity, a is acceleration and t is time.
Substitute each of the info given into the formula and calculate.
49 = 24 + (7)t
t = 3.57s
Answer:
15 Joules
Explanation:
work = charge x potential difference
= 10 x 1.5
= 15
