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irina [24]
3 years ago
12

Se tiene un objeto a una altura de 8 m generando 450 J de energía potencial cuál es la masa de dicho objeto

Physics
1 answer:
fgiga [73]3 years ago
8 0

Answer:

5.73kg

Explanation:

Given data

Height = 8m

Potential Energy = 450J

Required

The mass of the object

From the following expression

PE=mgh

we can substitute g= 9.81m/s^2 and the other parameters to get mass m

450= m*9.81*8

450= m* 78.48

m= 450/78.48

m= 5.73kg

Hence the mass is 5.73kg

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A man walks 30 m to the west, then 5 m to the east in 45 seconds.
katen-ka-za [31]

Answer:

a. Displacement=30²+5²=925= 30.4m

b. Total distance=30m+5m=35m

c. V=s/t. = 30.4/45=0.6m/s

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1 year ago
How does a generator work?
insens350 [35]

An electric generator is a device that converts mechanical energy obtained from an external source into electrical energy as the output.

It is important to understand that a generator does not actually ‘create’ electrical energy. Instead, it uses the mechanical energy supplied to it to force the movement of electric charges present in the wire of its windings through an external electric circuit. This flow of electric charges constitutes the output electric current supplied by the generator. This mechanism can be understood by considering the generator to be analogous to a water pump, which causes the flow of water but does not actually ‘create’ the water flowing through it.

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8 0
3 years ago
A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS
Flura [38]

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

[\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

(a) the electrical power generated for still summer day

T_s = T_{\infty} = 35 ^oC = 308 \ k

[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

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

(b)the electrical power generated for a breezy winter day

T_s = T_{\infty} = -15 ^oC = 258 \ k

[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

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

3 0
3 years ago
Science and math people where you at
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3 years ago
You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 _. If each light ha
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V = IR

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