Answer:
Explanation:
Astronauts face a wide range of dangers when maneuvering in space. Collecting rock samples requires breaking apart rocks on the moon's surface, which requires tools. These tools are sometimes small hammers or other times special drills. All of this process means moving and using force to break these rocks apart and every single movement can cause a mistake which can lead to a piece of the suit ripping or a failure in the machinery. Since there is no oxygen on the moon the tiniest failure in the suit can lead to the death of the astronaut. Therefore, having backup safety precautions, and quick repair scenarios is a must when collecting rock samples on the moon.
Answer:
140 beats per minute
Explanation:
There are 60 seconds in a minute and we know that Sherry felt 14 beats in 6 seconds if you multiply 6 by 10 you get 60 seconds which can be 1 minute. Then you multiply 14 by 10 since you 14 by 10 since you multiplied 6 by 10 and you get 140 beats per minute.
Answer:
I think its Pound or a gram
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA =
/ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper