The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.
![N_x=86.62N](https://tex.z-dn.net/?f=N_x%3D86.62N)
- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.
![N_y=F_V=mg-Tsin59\\](https://tex.z-dn.net/?f=N_y%3DF_V%3Dmg-Tsin59%5C%5C)
- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.
![F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N](https://tex.z-dn.net/?f=F_H%3DN_x%3DTcos59%5C%5C%5C%5CT%3D%5Cfrac%7BF_H%7D%7Bcos59%7D%20%3D%5Cfrac%7B86.62%7D%7B0.51%7D%3D%20169.84N)
- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,
![N_y= (40*9.8)-(169.8*sin59)=246.4N](https://tex.z-dn.net/?f=N_y%3D%20%2840%2A9.8%29-%28169.8%2Asin59%29%3D246.4N)
- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,
![N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N](https://tex.z-dn.net/?f=N%3D%5Csqrt%7BN_x%5E2%2BN_y%5E2%7D%20%3D%5Csqrt%7B%2886.62%29%5E2%2B%28246.4%29%5E2%7D%3D261.12N)
Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
Learn more about the tension here:
brainly.com/question/28106871
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Answer:
(a) the electrical power generated for still summer day is 1013.032 W
(b)the electrical power generated for a breezy winter day is 1270.763 W
Explanation:
Given;
Area of panel = 2 m × 4 m, = 8m²
solar flux GS = 700 W/m²
absorptivity of the panel, αS = 0.83
efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp
panel emissivity , ε = 0.90
Apply energy balance equation to determine he electrical power generated;
transferred energy + generated energy = 0
(radiation + convection) + generated energy = 0
![[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0](https://tex.z-dn.net/?f=%5B%5Calpha_sG_s-%5Cepsilon%20%5Calpha%28T_p%5E4-T_s%5E4%29%5D-h%28T_p-T_%5Cinfty%29%20-%20%5Ceta%20%5Calpha_s%20G_s%20%3D%200)
![[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s](https://tex.z-dn.net/?f=%5B%5Calpha_sG_s-%5Cepsilon%20%5Calpha%28T_p%5E4-T_s%5E4%29%5D-h%28T_p-T_%5Cinfty%29%20-%20%280.553-0.001T_p%29%5Calpha_s%20G_s)
(a) the electrical power generated for still summer day
![T_s = T_{\infty} = 35 ^oC = 308 \ k](https://tex.z-dn.net/?f=T_s%20%3D%20T_%7B%5Cinfty%7D%20%3D%2035%20%5EoC%20%3D%20308%20%5C%20k)
![[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k](https://tex.z-dn.net/?f=%5B0.83%2A700-0.9%2A5.67%2A10%5E%7B-8%7D%28T_p_1%5E4-308%5E4%29%5D-10%28T_p_1-308%29%20-%20%280.553-0.001T_p_1%290.83%2A700%20%3D%200%5C%5C%5C%5C3798.94-5.103%2A10%5E%7B-8%7DT_p_1%5E4%20-%209.419T_p_1%20%3D%200%5C%5C%5C%5CApply%20%5C%20%20%5C%20iteration%20%5C%20method%20%5C%20to%20%5C%20solve%20%5C%20for%20%5C%20T_p_1%5C%5C%5C%5CT_p_1%20%3D%20335.05%20%5C%20k)
![P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W](https://tex.z-dn.net/?f=P%20%3D%20%5Ceta%20%5Calpha_s%20G_s%20A%20%3D%20%280.553-0.001%20T_p_1%29%5Calpha_s%20G_s%20A%20%5C%5C%5C%5CP%20%3D%20%280.553-0.001%20%2A335.05%290.83%2A700%2A8%20%5C%5C%5C%5CP%20%3D%201013.032%20%5C%20W)
(b)the electrical power generated for a breezy winter day
![T_s = T_{\infty} = -15 ^oC = 258 \ k](https://tex.z-dn.net/?f=T_s%20%3D%20T_%7B%5Cinfty%7D%20%3D%20-15%20%5EoC%20%3D%20258%20%5C%20k)
![[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k](https://tex.z-dn.net/?f=%5B0.83%2A700-0.9%2A5.67%2A10%5E%7B-8%7D%28T_p_2%5E4-258%5E4%29%5D-10%28T_p_2-258%29%20-%20%280.553-0.001T_p_2%290.83%2A700%20%3D%200%5C%5C%5C%5C8225.81-5.103%2A10%5E%7B-8%7DT_p_2%5E4%20-%2029.419T_p_2%20%3D%200%5C%5C%5C%5CApply%20%5C%20%20%5C%20iteration%20%5C%20method%20%5C%20to%20%5C%20solve%20%5C%20for%20%5C%20T_p_2%5C%5C%5C%5CT_p_2%20%3D%20279.6%20%5C%20k)
![P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W](https://tex.z-dn.net/?f=P%20%3D%20%5Ceta%20%5Calpha_s%20G_s%20A%20%3D%20%280.553-0.001%20T_p_2%29%5Calpha_s%20G_s%20A%20%5C%5C%5C%5CP%20%3D%20%280.553-0.001%20%2A279.6%290.83%2A700%2A8%20%5C%5C%5C%5CP%20%3D%201270.763%20%5C%20W)
Answer:
Explanation: Decreasing in velocity
In an exothermic reaction, there is a transfer of energy to the surroundings in the form of heat energy. The surroundings of the reaction will experience an increase in temperature. Many types of chemical reactions are exothermic, including combustion reactions, respiration & neutralization reactions of bases & acids.
Metamorphic rock this possess often occurs in the mantle