Answer:
<em>2.753*10^-11N</em>
Explanation:
According to Newton's law of gravitation, the force between the masses is expressed as;
F = GMm/d²
M and m are the distances
d is the distance between the masses
Given
M = 3.71 x 10 kg
m = 1.88 x 10^4 kg
d = 1300m
G = 6.67 x 10-11 Nm²/kg
Substitute into the formula
F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²
F = 46.52*10^(-6)/1.69 * 10^6
F = 27.53 * 10^{-6-6}
F = 27.53*10^{-12}
F = 2.753*10^-11
<em>Hence the gravitational force between the asteroid is 2.753*10^-11N</em>
<em></em>
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 = / W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s
The correct answer for the mean would be 8
I hope that this helps !
I believe potential, to kinetic, to potential.
Answer
given,
Mass of the runner, M = 70 Kg
speed of the runner on the second base = 4.35 m/s
speed at the base = 0 m/s
Acceleration due to gravity,g = 9.8 m/s²
a) magnitude of mechanical energy lost
Mechanical energy lost is equal top gain in kinetic energy
b) Work done = Force x displacement
W = F. x
F = μ mg
W = μ mg . x
Work done is equal to 662.29 J
using the coefficient of the friction,μ = 0.7
x = 1.38 m
Hence, the runner will slide to 1.38 m.