I got you
Explanation:
normal force = 400 g cos 35
friction force up slope = .6 (400 g) cos 35
weight component down slope = 400 g sin 35
400 a = 400 g sin 35 - .6 (400 g cos 35)
a = g (sin 35 - .6 cos 35) = .082 g
I hope this helps you
Answer:
p = 8N/mm2
Explanation:
given data ;
diameter of cylinder = 150 mm
thickness of cylinder = 6 mm
maximum shear stress = 25 MPa
we know that
hoop stress is given as =
axial stress is given as =
maximum shear stress = (hoop stress - axial stress)/2
putting both stress value to get required pressure


t = 6 mm
d = 150 mm
therefore we have pressure
p = 8N/mm2
Answer:
The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.
Explanation:
The engine of the craft provides an upward thrust of
so that the space craft descends at a constant speed.
This implies that the net force on the space craft is zero.
The upward thrust will be equal to the downward gravitational pull by Callisto.
So the weight of the craft near the vicinity will be 3480 N.
I don't know this 1 I'm sorry I can't help you
Answer:
<em>The force required is 3,104 N</em>
Explanation:
<u>Force</u>
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = ma
Where a is the acceleration of the object.
On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

Where:
vf is the final speed
vo is the initial speed
a is the acceleration
t is the time
Solving for a:

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:


The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

F= 3,104 N
The force required is 3,104 N