Solar heating is the system composed of a fluid system to move the heat from the collector to its point of usage and a reservoir to stock the heat
<u>Explanation:</u>
The options given here like coal burning uses solid material as the source to heat and to generate energy. Similarly, nuclear power also requires solid particles like atoms or neutrons to strike the moderators forming energy.
In both of these cases, fluid system is present but it is used completely as coolant and to maintain the temperature. Thus, the remaining system that is solar heating has been done for water tanks where the fluid as water is used to move the heat from its collector to its point of usage. Even in solar system it is used as reservoir to stock the heat.
Answer:
![\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%20Acceleration%20%5C%20%28a%29%20%3D%202%20%5C%20m%2Fs%5E%7B2%7D%7D%20)
![\boxed{\sf Distance \ covered \ (s) = 100 \ m}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%20Distance%20%5C%20covered%20%5C%20%28s%29%20%3D%20100%20%5C%20m%7D%20)
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 20 m/s
Time taken (t) = 10 sec
To Find:
(i) Acceleration (a)
(ii) Distance covered (s)
Explanation:
![\sf (i) \ From \ 1^{st} \ equation \ of \ motion:](https://tex.z-dn.net/?f=%5Csf%20%28i%29%20%5C%20From%20%5C%201%5E%7Bst%7D%20%5C%20equation%20%5C%20of%20%5C%20motion%3A)
![\sf \implies v = u + at](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20v%20%3D%20u%20%2B%20at)
![\sf \implies 20 = 0 + a(10)](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%2020%20%3D%200%20%2B%20a%2810%29)
![\sf \implies 10a = 20](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%2010a%20%3D%2020)
![\sf \implies \frac{10a}{10} = \frac{20}{10}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20%5Cfrac%7B10a%7D%7B10%7D%20%20%3D%20%20%5Cfrac%7B20%7D%7B10%7D%20)
![\sf \implies a = 2 \: m/ {s}^{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20a%20%3D%202%20%5C%3A%20m%2F%20%7Bs%7D%5E%7B2%7D%20)
![\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:](https://tex.z-dn.net/?f=%5Csf%20%28ii%29%20%5C%20From%20%5C%202%5E%7Bnd%7D%20%5C%20equation%20%5C%20of%20%5C%20motion%3A)
![\sf \implies s = ut + \frac{1}{2} a {t}^{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20s%20%3D%20ut%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20a%20%7Bt%7D%5E%7B2%7D%20)
![\sf \implies s = (0)(10) + \frac{1}{2} \times 2 \times {(10)}^{2}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20s%20%3D%20%280%29%2810%29%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%202%20%5Ctimes%20%20%7B%2810%29%7D%5E%7B2%7D%20)
![\sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times {(10)}^{2}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20s%20%3D%20%20%5Cfrac%7B1%7D%7B%20%5Ccancel%7B2%7D%7D%20%20%5Ctimes%20%20%5Ccancel%7B2%7D%20%5Ctimes%20%20%7B%2810%29%7D%5E%7B2%7D%20)
![\sf \implies s = {10}^{2}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20s%20%3D%20%20%7B10%7D%5E%7B2%7D%20)
![\sf \implies s = 100 \: m](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20s%20%3D%20100%20%5C%3A%20m)
Answer:
I = 0.09[amp] or 90 [milliamps]
Explanation:
To solve this problem we must use ohm's law, which tells us that the voltage is equal to the product of the voltage by the current.
V = I*R
where:
V = voltage [V]
I = current [amp]
R = resistance [ohm]
Now, we replace the values of the first current into the equation
V = 180*10^-3 * R
V = 0.18*R (1)
Then we have that the resistance is doubled so we have this new equation:
V = I*(2R) (2)
The voltage remains constant therefore 1 and 2 are equals and we can obtain the current value.
V = V
0.18*R = I*2*R
I = 0.09[amp] or 90 [milliamps]
Answer:
48
Explanation:
you basically divide 1200 into 25
Answer:
The other angle is 120°.
Explanation:
Given that,
Angle = 60
Speed = 5.0
We need to calculate the range
Using formula of range
...(I)
The range for the other angle is
....(II)
Here, distance and speed are same
On comparing both range
![\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}](https://tex.z-dn.net/?f=%20%5Cdfrac%7Bv%5E2%5Csin%282%5Ctheta%29%7D%7Bg%7D%3D%5Cdfrac%7Bv%5E2%5Csin%282%28%5Calpha-%5Ctheta%29%29%7D%7Bg%7D)
![\sin(2\theta)=\sin(2\times(\alpha-\theta))](https://tex.z-dn.net/?f=%5Csin%282%5Ctheta%29%3D%5Csin%282%5Ctimes%28%5Calpha-%5Ctheta%29%29)
![\sin120=\sin2(\alpha-60)](https://tex.z-dn.net/?f=%5Csin120%3D%5Csin2%28%5Calpha-60%29)
![120=2\alpha-120](https://tex.z-dn.net/?f=120%3D2%5Calpha-120)
![\alpha=\dfrac{120+120}{2}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B120%2B120%7D%7B2%7D)
![\alpha=120^{\circ}](https://tex.z-dn.net/?f=%5Calpha%3D120%5E%7B%5Ccirc%7D)
Hence, The other angle is 120°