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gayaneshka [121]
3 years ago
11

Four round objects of equal mass and radius roll without slipping along a horizontal surface that then bends upward and backward

into an arc of half a circle. The objects all have the same linear speed initially. The objects are a hollow cylinder, a solid cylinder, a solid sphere, and a hollow sphere. The objects to go up the arc and exit the are going in the opposite direction they entered without falling off the arc. Now, several trials are run for each object. For each trial, the initial speed of the object is reduced until the object does not make it through the full arc. The speed needed for each object to just make it through the arc is recorded. Which of the following correctly lists the objects in order from fastest to slowest speed needed to make it through the arc?
A. Hollow cylinder, hollow sphere, solid cylinder, solid sphere
B. Hollow cylinder, solid sphere, solid cylinder, hollow sphere
C. Solid sphere, solid cylinder, hollow sphere, hollow cylinder
D. Solid sphere, hollow sphere, solid cylinder, hollow cylinder
E. Hollow sphere, solid cylinder, hollow cylinder, solid sphere
Physics
1 answer:
Reptile [31]3 years ago
5 0

Answer:

A

Explanation:

inertia Equation for All Object is k m R²

inertia Equation of Hollow Cylinder I = m R²

inertia Equation of Solid Cylinder I = ½ m R²

inertia Equation of Hollow Sphere I = ⅔ m R²

inertia Equation of Solid Sphere I = 2/5 m R²

to find the speed needed we can use Mechanical Energy (ME). or Total Energy.

ME = Translation Kinetic Energy + Rotation Kinetic Energi

ME = ½ m v² + ½ I w²

ME = ½ m v² (1 + k)

v² = 2ME / m(1+k)

sinnce all objects have the same mass, we can figure that v² is proportionally with 1/(1+k)

v²(hc) = 1/(1+1) = ½ = 0.5

v = 0.707

v²(sc) = 1/(1+½) = 2/3 = 0.67

v = 0.818

v²(hs) = 1/(1+⅔) = 3/5 = 0.6

v = 0.744

v²(ss) = 1/(1+²/5) = 5/7 = 0.71

v = 0.842

these is the ratio of speed they produce in the end of the arc with initial speed = 0 m/s

so, if we must give speed to the object so that it can reach the end of the arc, the fastest speed is given to hollow cylinder and the slowest speed is given to Solid Sphere

Then the answer is

HC, HS, SC, SS

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As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax
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Answer:

I=2.766\ kg.m^2

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3 0
4 years ago
An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
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Answer:

The density of this object is approximately 1.36\; {\rm kg \cdot L^{-1}}.

The density of the oil in this question is approximately 0.600\; {\rm kg \cdot L^{-1}}.

(Assumption: the gravitational field strength is g =9.806\; {\rm N \cdot kg^{-1}})

Explanation:

When the gravitational field strength is g, the weight (\text{weight}) of an object of mass m would be m\, g.

Conversely, if the weight of an object is (\text{weight}) in a gravitational field of strength g, the mass m of that object would be m = (\text{weight}) / g.

Assuming that g =9.806\; {\rm N \cdot kg^{-1}}. The mass of this 63.8\; {\rm N}-object would be:

\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}.

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was (63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}. Hence, the weight of water that this object displaced would be 47.0 \; {\rm N}.

The mass of water displaced would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}.

The volume of that much water (which this object had displaced) would be:

\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}.

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately 4.793\; {\rm L}.

The mass of this object is 6.50\; {\rm kg}. Hence, the density of this object would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately 4.793\; {\rm L}.

The weight of oil displaced would be equal to the magnitude of the buoyancy force: 63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}.

The mass of that much oil would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}.

Hence, the density of the oil in this question would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

7 0
2 years ago
A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume
Assoli18 [71]

Answer:

Hence the pressure is 3\times 10^5 Pa

Explanation:

Given data

Q=1500 J   system gains heat

ΔV=- 0.010 m^3     there is a decrease in volume

ΔU= 4500 J        internal energy decrease

We know work done is

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The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is 3\times 10^5 Pa

3 0
4 years ago
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