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3 years ago

PLEASEEEEE!!!!

Physics

1 answer:

Fudgin [204]3 years ago

3 0

Answer:

<h3>The answer is 2.0 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 100 g

volume = 50 mL

We have

$density = \frac{100}{50} = \frac{10}{5} \\$

We have the final answer as

Hope this helps you

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What are things that we can do to protect our climate for future generations?

Vilka [71]

<span>Reduce energy use.
Change the way you think about transportation. Walk or bike whenever possible.
Insulate your home. Insulate yourself and your home.
Make every drop count.
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2 years ago

A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which

julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = $11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}$

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = $390\ W/m.^{\circ}C$

Conduction of heat from the rod per second is given by:

$q = \frac{KA\Delta T}{L}$

where

$\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C$ = temperature difference between the two ends of the rod.

Thus

$q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s$

Now,

To calculate the mass, M of the ice melted per sec:

$M = \frac{q}{L_{w}}$

where

$L_{w}$ = Latent heat of fusion of water = 333 kJ/kg

$M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g$

5 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

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

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

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2 years ago

Which part of an atom is mostly empty space?

Veronika [31]

Answer:

The electron cloud is mostly empty space

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3 years ago

Astar begins as a blank a large cloud of gas and dust

crimeas [40]

This is a statement but yes a star forms inside nebulae which are gigantic clouds of gas. stars form inside as the gases own gravity pulls it together after which it becomes large enough to perform fusion and become a star.

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3 years ago