This question is not complete.
The complete question is as follows:
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?
Explanation:
a. Using the expression;
T = 2π√R/g
where R = radius of the space = diameter/2
R = 800/2 = 400m
g= acceleration due to gravity = 9.8m/s^2
1/T = number of revolutions per second
T = 2π√R/g
T = 2 x 3.14 x √400/9.8
T = 6.28 x 6.39 = 40.13
1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute
Answer:
Solution:
we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]
Position, velocity and acceleration are all based on the equation of motion.
The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.
x(t)=8sint
x'(t)=8cost
x"(t)=-8sint
now at time t=2pi/3,
position, x(t)=8sin(2pi/3)=4*squart(3)cm.
velocity, x'(t)=8cos(2pi/3)==4cm/s
acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2
so at present the direction is in y-axis.
Answer:

Explanation:
It is given that,
Depth of Death valley is 85 m below sea level, 
The summit of nearby Mt. Whitney has an elevation of 4420 m, 
Mass of the hiker, m = 65 kg
We need to find the change in potential energy. It is given by :



or

So, the change in potential energy of the hiker is
. Hence, this is the required solution.
Answer:
<em>The velocity after 12s is 50.4m/s</em>.
Explanation:
<em>In acceleration formula make velocity the </em><em>subject.</em>
<em> acceleration(a) = velocity(</em>v)÷time(t)
<h3><em> </em><em>velocity</em><em> </em><em>(</em><em>v)</em><em> </em><em>=</em><em> </em><em>acceleration</em><em>(</em><em>a)</em><em>×</em><em>t</em><em>ime</em><em>(</em><em>t)</em></h3>
<em>V </em><em>=</em><em> </em><em>4</em><em>.</em><em>2</em><em>m</em><em>/</em><em>s²</em><em>×</em><em>1</em><em>2</em><em>s</em>
<em>V </em><em>=</em><em> </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s</em>
<em>Therefore</em><em> the</em><em> </em><em>velocity</em><em> </em><em>after</em><em> </em><em>1</em><em>2</em><em>s</em><em> </em><em>is </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s.</em>