Answer:
La intensidad del campo eléctrico es 70312.5 
.
Explanation:
La perturbación que crea en torno a ella una carga eléctrica se representa mediante un vector denominado campo eléctrico.
Se dice que un campo eléctrico es uniforme en una región del espacio cuando la intensidad de dicho campo eléctrico es el mismo en todos los puntos de dicha región.
El campo eléctrico E creado por la carga puntual q en un punto cualquiera P se define como:
donde q es la carga creadora del campo, k es la constante electrostática y r es la distancia desde la carga fuente al punto P.
En este caso, los datos son:
- k= 9*10⁹
- q= 5*10⁻⁶ C
- r= 0.8 m
Reemplazando:
Resolviendo:
E= 70312.5
<em><u>La intensidad del campo eléctrico es 70312.5 </u></em><em><u>.</u></em>
Angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is 0.0466 rad/s.
Length of the cylinder, L = 9 mi
= (14.5mi)(1609.344 m / 1 mi)
= 23,335.48 m
Diameter of the cylinder, D = 4.78 mi
= (4.78 mi)(1609.344 m / 1 mi)
= 7692.645 m
Radius of the cylinder, r = D / 2
= ( 7692.645 m) / 2
= 3846.32 m
Centripetal acceleration is given by, ac = v^2 / r
We have the relation between linear velocity (v) and the angular velocity(ω) as
v = r ω
Then, ac = v2 / r
= (rω)2 / r
ac = ω^2 r
If the centripetal acceleration is equals the free fall acceleration of earth, then
ω^2 r = g
ω=0.0466 rad/s
- Angular acceleration is the term used to describe the rate of change in angular velocity. If the angular velocity is constant, the angular acceleration is constant.
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Answer:
B
Explanation:
The iris controls the amount of light that enters the eye by opening and closing the pupil. The iris uses muscles to change the size of the pupil. These muscles can control the amount of light entering the eye by making the pupil larger (dilated) or smaller (constricted).
Answer:
The amount of heat transfer is 21,000J .
Explanation:
The equation form of thermodynamics is,
ΔQ=ΔU+W
Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.
Substitute 0 J for W and 0 J for ΔU
ΔQ = 0J+0J
ΔQ = 0J
The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change
The heat transfer is,
ΔQ=Q (in
) −Q (out
)
Substitute 19000 J + 2000 J for Q(in) and 0 J for Q(out)
ΔQ=(19000J+2000J)−(0J)
=21,000J
Thus, the amount of heat transfer is 21,000J .
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
