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3 years ago

What is the mechanical advantage of the wheel and axle shown below?

Physics

2 answers:

Salsk061 [2.6K]3 years ago

7 0

Answer:

Mechanical advantage of the wheel and axle is 32.5

Explanation:

Given that,

The radius of the axle, $r_a=0.8\ cm$

The radius of wheel , $r_w=26\ cm$

We need to find the mechanical advantage of the wheel and axle. It can be calculated as :

$MA=\dfrac{r_w}{r_a}$

$MA=\dfrac{26}{0.8}$

MA = 32.5

So, the mechanical advantage of the wheel and axle shown below is 32.5 . Hence, this is the required solution.

BigorU [14]3 years ago

6 0

The correct answer is A. 32.5

Mechanical advantage is the ratio of force that is input into a machine to the force output.

Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.

MA=R/r where R is the radius of the wheel and r is the radius of the axle.

Substituting for the values in the question gives:

MA=26cm/0.8cm

=32.5

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Melanoma skin cancer

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3 years ago

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Which of the following illustrates an increase in potential energy? Group of answer choices a wind-up toy winding down a person

horrorfan [7]

Answer:

A person climbs a set of stairs

Explanation:

Potential energy is said to be possessed by an object due to its position. As the height from the ground level increase, the potential energy increases. It is calculated by the below formula as :

P = mgh

Out of the given options, the option that illustrates an increase in potential energy is option (b) i.e. a person climbs a set of stairs. As he steps one stair, its position from ground increases. It means its potential energy increases.

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3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

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3 years ago

A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C. How much water freezes onto the ice? The specific heat of

vladimir2022 [97]

For A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C, the amount of water that freezes onto the ice? is mathematically given as

x = 9.93 g

<h3>What is the amount of water that freezes onto the ice?</h3>

Where

Energy received = energy given out

Generally, the amount of water is mathematically given as

(53)(0.5)(30) = (80)(x)

Therefore

x = (49)(0.5)(16)/(80)

x = 9.93 g

In conclusion, the mass of water

x = 9.93 g