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Andrew [12]
3 years ago
8

Physics 1 Course. please help. ​

Physics
1 answer:
Aleks [24]3 years ago
6 0

Answer:

a. 1.64 m/s²

Explanation:

Centripetal acceleration is the square of tangential velocity divided by the radius.

a = v²/r

First, convert km/h to m/s.

30.0 km/h (1000 m/km) (1 h / 3600 s) = 8.33 m/s

Find the acceleration.

a = (8.33 m/s)² / (42.4 m)

a = 1.64 m/s²

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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
2 years ago
After a magnetic field is applied, how many energy levels would you expect for then=3,ℓ=2 state, including electron spin?
zloy xaker [14]

Answer:

The energy level is 5.

Explanation:

Given that,

n = 3

l = 2

We know that,

l shows the number of sub-shells and define the number of angular nodes.

n shows the number of electron shell.

m_{l} is a quantum number. It is define the number of energy level in a sub-shells .

m_{s} is define the spin of the electron.

So, The quantum number is

m_{l}=-2,-1,0,1,2

m_{s} is +\dfrac{1}{2} and -\dfrac{1}{2} for every energy level.

The energy level is 5.

Hence, The energy level is 5.

7 0
3 years ago
What is the benefit of developing the atomic mass unit as a standard unit of mass?
Alexeev081 [22]

Answer:

The atomic mass unit is 1/12 of an atom of carbon 12, and is a very small amount to represent in kilograms:

1m_{u}=1.66054x10^{-27}kg

m_{u} is atomic mass unit.

This is why the benefits of the atomic mass unit is that it makes the representation of atomic masses easier in terms of the simplicity of the numbers that are used to represent the masses. Also using the atomic mass unit it is easier to compare the masses of different atoms, These numbers would be very small and would require negative powers of 10 to represent them, so it is more convenient to use the atomic mass unit.

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2 years ago
Is the strange solid/liquid changing behavior of Oobleck the SAME as an ice cube melting?
Dmitrij [34]
I don’t think so because an ice cube melting needs heat and relies on temp while ooblecks transition from solid to quickly depends on force and speed
6 0
2 years ago
The electric field vector acts in the same direction as the force F vector if q is positive and in the opposite direction if q i
Contact [7]

Answer: Electric field vector is created by an electric force of a charged particle.

Explanation: The direction of the field vector can be determined by applying Coulomb’s Law which explains the electric force between charge particles. If q is positive the force is repulsive, to a test charge ( which is positive always ) and if q is negative the force is attractive to a test charge.

5 0
2 years ago
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