Assuming the force the car exerts is 140,000 N and that the car is stationary on a flat surface. You can apply Newton's Second Law of Motion.
F = MA
Where the acceleration is gravity (-9.8m/s).
140,000 = M*-9.8
M = 14285.71kg
<span>You can prove that moon is in motion by taking two photos with the following features:
They are taken at the same hour and minute.
They are taken on two different nights.
They are taken from the same location.
This would allow an accurate comparison and would should the moon's motion.
I hope this helped! :)</span>
Answer:
a)F=(208 i+156 j) kN
b)F+W=219 kN
Explanation:
Solution:
To determine the components of F
F=(260 kN).()
=(208 i+156 j) kN
F=(208 i+156 j) kN
b)weight W expressed in component and add into the
Force F.
W=-(8800 kg).(9.81 m/s^2)j
F+W=(208 i+[156-86.3]j) kN
F+W=
=219 kN
F+W=219 kN
Answer:
1) duration of the sessions
2) daily pattern days and ..
Answer:
a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36
Explanation:
a) This is a balancing act.
Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive
-w x - W x₂ + R y = 0 (1)
usemso trigonometry to find distances
cos 60.08 = x / 7.5
x = 7.5 cos 60.08
x = 3.74 m
fireman
cos 60.08 = x₂ / 4
x2 = 4 cos 60
x2 = 2 m
wall support
sin 60.08 = y / 15
y = 15 are 60.08
y = 13 m
we substitute in equation 1
R y = w x + W x2
R = (w x + W x2) / y
R = (500 3.74 +800 2) / 13
R = 266.92 N
now let's write the expressions for the translational equilibrium
X axis
R -fr = 0
R = fr
fr = 266.92 N
Y Axis
Fy - w-W = 0
fy = 500 + 800
fy = 1300 N
b) ask the friction coefficient
the firefighter's distance is
cos 60.08 = x₃ / 9.00
x₃ = 9 cos 60
x₃ = 5.28 m
from equation 1
R = (w x + W x₃) / y
R = 500 3.74 + 800 5.28) / 13
R = 468.769 N
we saw that
fr = R = 468.769
The expression for the friction force is
fr = μ N
in this case the normal is the ratio to pesos
N = Fy
N = 1300 N
μ N = fr
μ = fr / N
μ = 468,769 / 1300
μ = 0.36