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Oxana [17]
2 years ago
7

If a star were located exactly at each celestial pole, the corrected altitude of the star would equal __________.

Physics
1 answer:
dimulka [17.4K]2 years ago
4 0

Answer:

the observer's latitude

Explanation:

the observer's latitude

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De manera individual subraya la respuesta correcta: A) El Método Científico de la física experimental y su búsqueda de respuesta
alexira [117]

Answer:

a. La ciencia.

Explicación:

El método científico de la física experimental y su búsqueda de respuestas tiene un principio básico que es la ciencia. La ciencia es la aplicación del conocimiento del mundo natural y social siguiendo una metodología sistemática basada en evidencias. Sin evidencias no hay aceptación de la observación ni se debe hacer teoría porque la evidencia es el principal criterio para aceptar o rechazar una hipótesis.

6 0
3 years ago
A machine has a mechanical advantage of 4.5. What force is put out by the machine if the force applied to the machine is 800 N?
DerKrebs [107]
<span>Describe the action of protein carrier: ATP synthase? Explain why it is important to have build up of Hydrogen ions inside the thylakoid? 

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5 0
3 years ago
A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when
algol [13]
I would say 648858. bc yes
4 0
3 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0
3 years ago
A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
notsponge [240]

Answer:

Q = \frac{0.068}{E}

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

QE = F_x

mg = F_y

since string makes 30 degree angle with the vertical so we will have

tan\theta = \frac{F_x}{F_y}

tan30 = \frac{QE}{mg}

Q = \frac{mg}{E}tan30

Q = \frac{0.012\times 9.81}{E} tan30

Q = \frac{0.068}{E}

where E = electric field intensity

5 0
3 years ago
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