You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
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Answer:
Explanation:
We are given that
Linear charge density of wire=
Radius of hollow cylinder=R
Net linear charge density of cylinder=
We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R
By Gauss theorem
Where surface area of cylinder=
Answer:
T₂ = 5646 K
Explanation:
Let's start by finding the power received by the first disc, for this we use Stefan's law
P = σ. A e T⁴
Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of the disk, T the absolute temperature and e the emissivity that for a black body is 1
The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is
I = P / A
Writing this expression for both discs
I₁ A₁ = I₂ A₂
I₂ = I₁ A₁ / A₂
The area of a sphere is
A = 4π r²
I₂ = I₁ (r₁ / r₂)²
r₂ = r₁ ± 5
I₁ = I₂ ( (r₁ ± 5)/r₁)²
.
Let's write the Stefan equation
P / A = σ e T⁴
I = σ e T⁴
This is the intensity that affects the disk, substitute in the intensity equation
σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²
The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white
e₁ T₁⁴ = T₂⁴ (r1 + 5)²/r₁²
T₁ = T₂ {(e₂/e₁)}^{1/4} √(1 ± 1/ r₁)
If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term
(1 ±x) n = 1 ± n x
Where x = 5 / r₁ << 1
We replace
T₁ = T₂ {(e₂/e₁)}^{1/4} (1 ± ½ 5/r1)
T₁ = T₂ {(e₂)}^{1/4} (1 ± 5/2 1/r1)
If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach
T₁ = T₂ {(e₂)}^{1/4} ¼
T<u>₁</u> = T₂ 0.9
5500 = T₂ 0.974
T₂ = 5646 K
A) Zero
The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.
B) -9.8 m/s^2
The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:
where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:
where the negative sign means it points downward.
C) 7.6 m/s
The x-component of the shot's velocity is given by:
where
is the initial velocity
is the angle of the shot
Substituting into the equation, we find
D) 9.3 m/s
The y-component of the shot's velocity is given by:
where
is the initial velocity
is the angle of the shot
Substituting into the equation, we find
E) 7.6 m/s
We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:
F) -11.1 m/s
The y-component of the velocity at time t is given by:
where
is the initial y-velocity
a = g = -9.8 m/s^2 is the vertical acceleration
t is the time
Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:
where the negative sign means the vertical velocity is now downward.