If the first stage provides a thrust of 5.31 mega-newtons [MN] and the space shuttle has a mass of 5,760,000 pound-mass [lbm
1 answer:
Answer:
The acceleration of the spacecraft is 16,367.62 miles/hr²
Explanation:
Given;
thrust of the spacecraft, F = 5.31 x 10⁶ N
mass of the spacecraft, m = 5,760,000 lbm = 2612692.051 kg
The acceleration of the spacecraft in m/s² is given by Newton's second law of motion;
F = ma
a = F / m
The acceleration in miles per hour square is given by;
Therefore, the acceleration of the spacecraft is 16,367.62 miles/hr²
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Answer:
y = 67.6 feet, y = 114.4/ (22 - 3t)
Explanation:
For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram
Large triangle Projector up to the screen
tan θ = y / L
For the small triangle. Projector up to the person
tan θ = y₀ / (L-d)
The angle is the same, so we equate the two equations
y₀ / (L -d) = y / L
y = y₀ L / (L-d)
The distance from the screen (d), we look for it with kinematics
v = d / t
d = v t
we replace
y = y₀ L / (L - v t)
y = 5.2 22 / (22 - 3 t)
y = 114.4 (22 - 3t)⁻¹
This is the equation of the shadow height change as a function of time
For the suggested distance the shadow has a height of
y = 114.4 / (22-13)
y = 67.6 feet
Answer:
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
Explanation:
As we know that the relation between electric field and electric potential is given as
here if we say that potential is constant because electric field sensor is moving along equi-potential line.
Then we will say
V = constant
so we have
so electric field will remain constant always in magnitude and always remains perpendicular to the surface
so we have
b. Constant magnitude, but varying direction, perpendicular to the equipotential.