A woman pushes a 3.2kg textbook on a table as she walks forward 3.5 meters. How much work does the vertical component of the for
1 answer:
Answer:
The done by the vertical component of force is zero.
Explanation:
Given data,
The mass of the textbook the women pushes horizontally, m = 3.2 kg
The displacement of the textbook, S = 3.5 m
Let the force of the women acts on the book horizontally,
Therefore, the horizontal component of force is maximum and the vertical component is zero.
If F is the force applied by the women, then the horizontal and vertical component of the force is,
Since the force is acting along with the horizontal x component, the vertical component of the force is zero.
Hence, the done by the vertical component of force is zero.
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Efficiency is calculated through dividing the actual mechanical advantage by the hypothetical mechanical advantage:
- the actual mechanical advantage is 9J because that's how much work the light bulb doing
- the hypo. mechanical advantage is 100J. Ideally, in a perfect world, the light bulb can convert 100J input into 100J output, but do to resistance and other factors it is not possible.
change the decimal to a percentage:
the light bulb had 9% efficiency
- the actual mechanical advantage is 9J because that's how much work the light bulb doing
- the hypo. mechanical advantage is 100J. Ideally, in a perfect world, the light bulb can convert 100J input into 100J output, but do to resistance and other factors it is not possible.
change the decimal to a percentage:
the light bulb had 9% efficiency
Answer:
≈ 2.1 R
Explanation:
The moment of inertia of the bodies can be calculated by the equation
I = ∫ r² dm
For bodies with symmetry this tabulated, the moment of inertia of the center of mass
Sphere = 2/5 M R²
Spherical shell = 2/3 M R²
The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass
I = + M D²
Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation
Let's start with the spherical shell, axis is along a diameter
D = 2R
Ic = + M D²
Ic = 2/3 MR² + M (2R)²
Ic = M R² (2/3 + 4)
Ic = 14/3 M R²
The sphere
Is = + M [
²
Is = Ic
2/5 MR² + M ² = 14/3 MR²
² = R² (14/3 - 2/5)
= √ (R² (64/15)
= 2,066 R