convection
please mark brainliest any other problems or questions feel free to ask
Answer:
Option e) 320 s
Explanation:
Here, distance = 3.0 km = 3000 m
The velocity of boat when it is going upstream;
Upstream velocity = velocity of boat in still water - velocity of river flow
So, Upstream velocity 
So,Time to go upstream
The velocity of boat when it is going downstream;
Downstream velocity = velocity of boat in still water + velocity of river flow
So, Downstream velocity 
So,Time to go downstream
So, total time (t) = 
Option E is the correct answer.
Answer:
2.464 cm above the water surface
Explanation:
Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.
We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):
weight of the block = 0.78 * 11.2^3 gr
Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.
So the weight of the volume of water displaced is:
weight of water = 1 * 11.2^2 * x
we make both weight expressions equal each other for the floating requirement:
0.78 * 11.2^3 = 11.2^2 * x
then x = 0.78 * 11.2 cm = 8.736 cm
This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:
11.2 cm - 8.736 cm = 2.464 cm
Answer:
Explanation:
<u>Given Data:</u>
Length = l = 820 mm = 0.82 m
Acceleration due to gravity = g = 9.8 ms⁻²
<u>Required:</u>
Frequency = f = ?
<u>Formula:</u>
<u>Solution:</u>
![\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%20%3D%5Cfrac%7B1%7D%7B2%20%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7Bg%7D%7Bl%7D%20%7D%20%5C%5C%5C%5CPut%5C%20the%5C%20givens%5C%5C%5C%5Cf%3D%5Cfrac%7B1%7D%7B2%20%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7B9.8%7D%7B0.82%7D%20%7D%5C%5C%5C%5C%20f%20%3D%200.159%20%5Ctimes%20%5Csqrt%7B11.95%7D%20%5C%5C%5C%5Cf%3D0.159%20%5Ctimes%203.457%5C%5C%5C%5Cf%3D0.55%20%5C%20Hz%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Answer:
E = 10t^2e^-10t Joules
Explanation:
Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.
The energy E stored in the inductor can be expressed as
E = 1/2Ll^2
Substitutes the inductor L and the current I into the formula
E = 1/2 × 0.2 × ( 10te^-5t )^2
E = 0.1 × 100t^2e^-10t
E = 10t^2e^-10t Joules
Therefore, the energy stored in the inductor is 10t^2e^-10t Joules
