Answer:
G.P.E = 368.3
Explanation:
Given the following data;
Mass = 2.63kg
Height, h = 14.29m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the gravitational potential energy;
Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.
Mathematically, gravitational potential energy is given by the formula;
Where;
G.P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
G.P.E = 368.3
Note: the unit of gravitational potential energy is Joules.
<span>(a).the heat trasfer surface area and heat flux on the surface of filament are
Area of Surface= µDL=3.14(0.05cm)(5cm)= 0.785 cm square
qs=Q/Area of surface= 150W/0.785= 191W/cmsq.=1.91x10Âłx10ÂłW/Msq
(b). the heat surface on the surface of heat bulb
Area of surface = 3.14xD²= 3.14(8CM)²= 201.1cm²
qs=Q/Area of surface=150w/201.1cm²=0.75 w/cm²= 7500w/m²
the amount and cost of electrical energy consumed during one period is
Electrical Consumption=QΛt=(0.15 KW)(365X8h/yr)=438 k Wh/yr
Annual cost= 438 kWh/yr)($.08/kWh)= $ 35.04 /yr</span>
The diameter of the hose is 6.34 cm.
<em>"Your question is not complete, it seems to be missing the following information";</em>
the flow rate of water in the pipe is 0.012 m³/s
The given parameters;
- velocity of water in the hose, v = 3.8 m/s
- flow rate of water in the hose, Q = 0.012 m³/s
Volumetric flow rate is directly proportional to the product of the area of the hose through which the water flows and the velocity of the water flowing through the hose.
Q = Av
where;
<em>Q is the volumetric flow rate</em>
<em>A is the area of the hose</em>
<em>v is the velocity of flow</em>
The area of the hose is calculated as follow;
The diameter of the hose is calculated as follows;
Thus, the diameter of the hose is 6.34 cm.
Learn more here: brainly.com/question/15061170
They travel the slowest through gasses
Looking for t, the computation would be:
t = 2[initial vertical velocity] / g = 2u/g
=> 9 = 2u/9.81
or u = 9 x 9.81 / 2 = 44.16 m/s (upwards)
conservation of energy: loss in potential energy(mgh) would be equal to gain in kinetic energy (½mv²)
=> ½mv² - ½mu² = mg29
=> v² = 58(9.81) + 44.16^2
= 568.98 + 1950.1056 = 2519.0856
getting the square root will give us the speed of the rock, which is:
=> v = √[2519.0856] ~= 50.19 m/s (directed downwards)
