In ptolemy’s earth-centered model for the solar system, venus always stays close to the sun in the sky and, because it always st
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The gravitational potential energy is 25.6 J
Explanation:
The gravitational potential energy (GPE) of an object is given by:
where
m is the mass of the object
g is the gravitational field strength
h is the height of the object above the ground
In this problem, we have
m = 8 kg is the mass of the brick
g = 1.6 N/kg is the gravitational field strength on the moon
h = 2 m is the height of the brick above the ground
Substituting,
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Answer:
a) The module's acceleration in a vertical takeoff from the Moon will be
b) Then we can say that a thrust of won't be able to lift off the module from the Earth because it's smaller than the module's weight (
).
Explanation:
a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)
Where:
thrust
module's mass
moon's gravity acceleration
module's acceleration during takeoff
Then, we can find the acceleration like this:
The module's acceleration in a vertical takeoff from the Moon will be
b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.
Then we can say that a thrust of won't be able to lift off the module from the Earth because it's smaller than the module's weight (
).
Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
new velocity coming out of the hose then is
44 ft/sec
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
new velocity coming out of the hose then is
44 ft/sec
Answer:
I₂ = 0.04 W / m²
Explanation:
Sound intensity is the power emitted between the unit area
I = W / A
W = I A
sound is a wave that travels in space whereby its energy is distributed on the surface of a sphere
A = 4π r²
we substitute
W = I (4π r²)
the emission power is constant, so the intensity at two different points is
W = I₁ 4π r₁² = I₂ 4π r₂²
so the equation is
I₁ r₁² = I₂ r₂²
In this case the units are not shown in the exercise, suppose that all units are in the SI system
I₂ =
let's calculate
I₂ = 4
I₂ = 0.04 W / m²