Explanation:
Expression to calculate thermal resistance for iron (
) is as follows.
where, 
= length of the iron bar
= thermal conductivity of iron
= Area of cross-section for the iron bar
Thermal resistance for copper (
) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]
where, 
= length of copper bar
= thermal conductivity of copper
= Area of cross-section for the copper bar
Now, expression for the transfer of heat per unit cell is as follows.
Q = 
Putting the given values into the above formula as follows.
Q = 
= 
= 2.92 Joule
It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,
P = 
Here, T is 1 second so, power conducted is equal to heat transferred.
So, P = 2.92 watt
Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.
<span>Assuming continuous operation (24/7), we can say that
Energy produced : Energy per hour * 24 (number of hours in a day) - 365 (number of days in a year.
Energy per hour: 2050 * 1.055 = 2162.75 kg.
So, we proceed to calculate the results
E: 2162.75 * 24 * 365 = 18,945,690 kj per year.
Now, we transform kj to megajoule, remembering that kilo is 10*3 and mega is 1'*6, so we divide the result by 1,000 in order to get the results in megajoules, and the answer would be:
18,945.69 megajoules can be produced per year.</span>
The period of a wave is the time it takes the wave to complete one cycle (at a fixed location).
So if a wave completes one cycle in <em>2 seconds</em>, then that is its period.
An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)
The stone will take 2.89 seconds to hit the water.
The time required by the stone to hit the water is calculated by the second equaiton of motion
s=ut+
gt^2
41=0×t+×9.81×t²
t=2.89 seconds
