Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts. The supply storage area of the lunar outpost, where gravity is 1.63 m/s2, can only support 1 x 10^5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost
Answer:
601,220N
Explanation:
Given that:
Gravity at lunar outpost = 1.63m/s²
Acceleration due to gravity on earth = 9.8m/s²
Supported weight = 1 * 10^5 N
Maximum weight of supplies as measured on earth;
(Ratio of the gravities) * weight of supplies
(9.8m/s² / 1.63m/s²) * (1 * 10^5 N)
6.0122 * (1 * 10^5)
6.0122 * 10^5 N
= 601,220 N
Answer:
E) 0 J
Explanation:
Hi there!
The net work done on the elevator can be calculated by adding the work done by the forces acting on the elevator. In this case, we can distinguish two forces acting in the vertical direction, the force that elevates the elevator (in the upward direction) and the gravity force (in the downward direction). Using Newton´s second law in the vertical direction:
F - Fg = m · a
Where:
F = force of the elevator.
Fg = force of gravity.
m = mass of the elevator.
a = acceleration.
Since the elevator is moving at constant velocity, the acceleration is zero. Then:
F - Fg = 0
F = Fg
Now, let´s calculate the work done by these forces during 20 m. The equation of work is the following:
W = F · d
Where:
W = work.
F = applied force.
d = distance.
The net work done on the elevator can be expressed as the work done by the elevator plus the work done by gravity:
Wnet = W elevator + W gravity
Notice that gravity does work in the direction opposite to the movement of the elevator, then, it has to be negative:
Wnet = W elevator - W gravity
Wnet = F · d - Fg · d
Wnet = (F - Fg) · d
Since F - Fg = 0
Then:
Wnet = (F - Fg) · d = 0 N · d = 0 J
The correct answer is E) 0 J.
The awnser is Plastic print
Answer:
x=502.837m
v=61m/S
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X (2)
X=Xo+ VoT+0.5at^{2} (3)
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for the part a we can use the ecuation number 2
a=3.7m/s^2
Vf=61m/s
Vo=0m/s
for the part b
the problem indicates that the speed with which it touches the ground is 61m / s
1 the angle of incidence is always the angle that comes into the mirroe
