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Marta_Voda [28]
3 years ago
11

hich one of the following statements could be an operational definition of electric current?View Available Hint(s)Which one of t

he following statements could be an operational definition of electric current?The magnitude of the physical quantity of electric current I in a wire equals the magnitude of the electric charge q that passes through a cross section of the wire divided by the time interval Δt needed for that charge to pass.
Physics
1 answer:
klemol [59]3 years ago
8 0

Answer:

it is True as the operational definition of electric current.

Explanation:

The definition of electric current is

         I = dQ / dt

By convention the direction of the current is the direction in which a positive charge flows.

The initial expression is the derivative that is the change of the load in the unit of time and this occurs in a given cross-sectional cable.

The proposed definition is the same as this, so it is True as the operational definition of electric current.

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Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de
marshall27 [118]
Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos: 

Altura o distancia recorrida: 40 m
Vo: 6 m/s 
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes. 

Entonces tenemos que:

Vf ^{2} -Vo ^{2} =2 x g x h

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2}  \\  \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2}  }  \\ Vf=28,64 m/s

Que tengas un buen día!
6 0
4 years ago
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
3 years ago
A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show
Lady bird [3.3K]

Answer:

change in height  is 1.664 mm

Explanation:

Given data

drops = 3.00 mm

diameter = 5.00 cm = 0.05 mm

decrease  =  350 cm^3

temperature = 95°C to 44.0°C

to find out

the decrease in millimeters in level

solution

we  will calculate here change in volume so we can find how much level is decrease

change in volume = β v change in temp   ...............1

here change in volume = area× height

so = \pi /4 × d² h

so we can say change in volume =  \pi /4 × d² × change in height   .......2

so from equation 1 and 2  we calculate change in height

( β(w) -β(g) )× v× change in temp = \pi /4 × d² × change in height

change in height = 4 × ( β(w) -β(g) ) v× change in temp / \pi /4 × d²

put all value here

change in height = 4 × ( 210 - 27 )(350 ) 10^{-6}× (95-44) / \pi /4 × 0.05²

change in height  is 1.664 mm

5 0
3 years ago
Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
kifflom [539]

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

     

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = \sqrt[4]{P/Ae}

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

4 0
3 years ago
Which organelle is the powerhouse of the cell, the site of cellular respiration? A) 2 - nucleus B) 5 - endoplasmic reticulum C)
Paladinen [302]
D) 9- mitochondria
Hope this helps :D
7 0
3 years ago
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