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Rina8888 [55]
3 years ago
13

How is the ideal mechanical advantage of a wheel and axle calculated?

Physics
2 answers:
kvasek [131]3 years ago
8 0
Mechanical Advantage is the value which tells us the factor by which the machine is able to convert the force applied to it to its output.

Ideal Mechanical Advantage of wheel and axle is calculated by 

IMA = Wheel Radius / Axle Radius

By virtue, the higher the computed IMA the lower the force needed to be exerted, this is the purpose for using machine, lower input force but larger output. (The smalled the axle radius compared to the wheel radius is the better setup.)
maria [59]3 years ago
7 0

Divide the radius of the wheel by the radius of the axle

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The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part
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Answer:

B. distance/potential

Explanation:

Quizlet

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3 years ago
What is the relationship between mass gravity and weight?
Basile [38]
Fg=m•g || IE: Weight = mass x gravity
Therefore, the relationship are as follows:
mass and gravity are inversely proportional 
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weight and gravity are directly proportional 
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Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i
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Answer:

The final speed for the mass 2m is v_{2y}=-1,51\ v_{i} and the final speed for the mass 9m is v_{1f} =0,85\ v_{i}.

The angle at which the particle 9m is scattered is \theta = -66,68^{o} with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

\vec{p}=m\vec{v}

p_{xi} =p_{xf}

In the x axis before the collision we have

p_{xi}=9m\ v_{i} - 2m\ v_{i}

and after the collision we have that

p_{xf} =9m\ v_{1x}

In the y axis before the collision p_{yi} =0

after the collision we have that

p_{yf} =9m\ v_{1y} - 2m\ v_{2y}

so

p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}

then

p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}

<u>Conservation of kinetic energy:</u>

\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}

so

\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}

Putting in one side of the equation each speed we get

\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the  speed of the 9m particle:

v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}

the magnitude of the final speed of the particle 9m is

v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }

v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}

The tangent that the speed of the particle 9m makes with the -y axis is

tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}

As a vector the speed of the particle 9m is:

\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}

As a vector the speed of the particle 2m is:

\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}

8 0
3 years ago
An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the ob
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Answer:

the force of attraction between the two charges is 3.55 N.

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first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N

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2 years ago
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Ksju [112]
Part 1) Which metal will cool the fastest?
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\frac{\Delta Q}{\Delta t}  = -k  \frac{A \Delta T}{x}
where:
\Delta Q is the heat exchanged
\Delta t is the time interval
k is thermal conductivity of the material
A the  surface where the exchange of heat occurs
\Delta T the variation of temperature
x is the thickness of the material
We see that the heat flow rate \frac{\Delta Q}{\Delta t} is linearly proportional to k, the thermal conductivity of the material. So, the larger k, the fastest the metal will cool. 
If we have a look at the thermal conductivity of each metal, we find:
- Aluminium: 237 W/(mK)
- Copper: 401 W/(mK)
- Gold: 314 W/(mK)
- Platinum: 69 W/(mK)
Therefore, copper is the material with highest heat flow rate, so the metal which cools fastest.

Part 2) Which sample of copper demonstrates the greatest increase in temperature
To solve this part, we can have a look at how the amount of heat exchanged Q is related to the increase in temperature \Delta T:
Q=m C_S \Delta T
where m is the mass and Cs the specific heat of the material. Re-arranging the formula, we get
\Delta T= \frac{Q}{m C_s}
therefore, we see that the increase in temperature is inversely proportional to the mass m. This means that the block that will show the largest increase in temperature is the block with the smallest mass, so the correct answer is A) 0.5 kg.
4 0
3 years ago
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