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1 year ago

If a current-carrying wire is in a magnetic field, in what direction will a force be exerted on the wire?.

Physics

1 answer:

astraxan [27]1 year ago

4 0

Answer:

The magnetic force on a current-carrying wire is perpendicular to both the wire and the magnetic field with direction given by the right hand rule.

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If the mass of material is 44 grams and the volume of the material is 8cm^3 what would the density of the material be?

Alenkasestr [34]

Density = mass ÷ volume

D= 44g ÷ 8 cm^3

D = 5,5 (round it) 6 g/cm^3

6 0

2 years ago

The variation in the pressure of helium gas, measured from its equilibrium value, is given by ΔP = 2.9 × 10−5 cos (6.20x − 3 000

nadya68 [22]

Answer:

The wavelength of this wave is 1.01 meters.

Explanation:

The variation in the pressure of helium gas, measured from its equilibrium value, is given by :

$\Delta P=2.9\times 10^{-5}\ cos(6.2x-3000t)$ ..............(1)

The general equation is given by :

$\Delat P=P_o\ cos(kx-\omega t)$ ...........(2)

On comparing equation (1) and (2) :

$k=6.2$

Since, $k=\dfrac{2\pi}{\lambda}$

$\dfrac{2\pi}{\lambda}=6.2$

$\lambda=1.01\ m$

So, the wavelength of this wave is 1.01 meters. Hence, this is the required solution.

6 0

2 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

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

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