ThIs is the same type of problem
find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given 5 m
thus speed ,= distance/time
speed = 5/0.77
= 6.45 m/s
Answer: please see below
Explanation:
A manned space exploration is defined as the exploration of individuals --- astronauts in space using a spacecraft as a vehicle and are responsible for operating its controls
The extreme conditions in space that challenge manned space exploration is as follows.
1. extreme loud sound waves cause by the launch of spacecraft which can shatter the spacecraft
2. extreme Temperatures in space ranging from extreme hot temperatures (near the sun) to extreme cold temperatures ( below freezing point out of space.
3.micrometeorite showers responsible for sandblasting can damage spacecraft.
4.Ultra violet Radiation which can alter the control unit of the spacecraft
Because of theses extreme conditions that pose challenges to space explorations, necessary precautions should be taken into consideration to be able to overcome such challenges. These precautions include building the spacecraft and the control unit in such a way that can resist these harmful conditions, also taking in mind safe escape routes for the astronauts in case of failures.
Answer:
Explanation:
Work in pumping water from the tank is given as
W = ∫ y dF. From a to b
Where dF is the differential weight of the thin layer of liquid in the tank, y is the height of the differential layer
a is the lower limit of the height
b is the upper limit of the height.
We know that, .
F = ρVg
Where F is the weight
ρ is the density of water
V is the volume of water in tank
g is the acceleration due to gravity
Then,
dF = ρg ( Ady)
We know that the density and the acceleration due to gravity is constant, also the base area of the tank is constant, only the height that changes.
Then,
ρg = 62.4 lbs/ft³
Area = L×B = 3 × 9 = 27ft²
dF = ρg ( Ady)
dF = 1684.8dy
The height reduces from 12ft to 0ft
Then,
W = ∫ y dF. From a to b
W = ∫ 1684.8y dy From 0 to 12
W = 1684.8y²/2 from 0 to 12
W = 842.4 [y²] from y = 0 to y = 12
W = 842.4 (12²-0²)
W = 121,305.6 lb-ft
Answer: a) B = 6811N
b) m = 603.2kg
c) 86.8%
Explanation: <em>Buoyant force</em> is a force a fluid exerts on a submerged object.
It can be calculated as:
where:
is density of the fluid the object is in;
is volume of the object;
g is acceleration due to gravity, is constant and equals 9.8m/s²
a) For the hollow plastic sphere, density of water is 1000kg/m³:
B = 6811N
b) Anchored to the bottom, the forces acting on the sphere are <u>Buoyant</u>, <u>Tension</u> and <u>Force due to gravity</u>:
B = T + 
B = T + mg
mg = B - T
Calculating:
m = 603.2kg
c) When the shpere comes to rest on the surface of the water, there are only <u>buoyant</u> <u>and</u> <u>gravity</u> acting on it:
B = m.g
= 0.6032m³
Fraction of the submerged volume is:
= 
= 0.868
<u />
Answer:
24.3KW
Explanation:
A)The kinetic energy is changing, the potential energy is changing and the chemical energy in form of fuel powering the engine also is changing
The kinetic energy is increasing as the body gain speed, the potential energy also increases as the body gain height against gravity and the chemical energy in form of fuel decreases as the body burn the fuel to create a lifting force
B) The workdone by the lifting force = the change in kinetic energy + the change in potential energy
C)The time taken in seconds to do the work is the variable needed
D) average power generated by the lifting force = (change in kinetic energy + change in potential energy) / time taken in seconds
Average power = 1/2 * m(mass) (Vf-Vi)^2 + mg(hf-hi) /t where vf is final speed and vi is initial speed at rest = 0, similarly, hf = final height and hi = initial height.
Average power = 1/2*810*7^2 + 810*9.81*8.2/3.5s
Average power = (19845+65158.02)/3.5 = 24286.577 approx 24.3kW
