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3 years ago

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A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid

e and brings it up to its proper turning rate of one complete rotation every 5.5 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding?

1 answer:

tatyana61 [14]3 years ago

6 0

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

coefficient of friction (Uk) = ?

we can get the minimum coefficient of friction from the equation below

centrifugal force = frictional force

m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = $\frac{ r x ω^{2} }{g}$

where ω (angular velocity) = $\frac{2π}{time}$

= $\frac{2π }{5.5}$ = 1.14

Uk = $\frac{ 5.7 x 1.14^{2} }{9.8}$ = 0.76

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3 years ago

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If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

3 years ago

Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°

lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía ( $S$ ), medida en joules por Kelvin, tenemos la siguiente expresión:

$dS = \frac{\delta Q}{T}$ (1)

Donde:

$Q$ - Ganancia de calor, en joules.

$T$ - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

$dS = \frac{L_{v}}{T}\cdot dm$ (1b)

Donde:

$m$ - Masa del sistema, en kilogramos.

$L_{v}$ - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

$\Delta S = \frac{\Delta m \cdot L_{v}}{T}$

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que $m = 0.50\,kg$ , $L_{v} = 2.7\times 10^{5}\,\frac{J}{kg}$ and $T = 630.15\,K$ , entonces el cambio de entropía es:

$\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}$

$\Delta S = 214.235 \,\frac{J}{K}$

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

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2 years ago

An instrument is defective for all of the following conditions EXCEPT:_______

NARA [144]

Answer:

I believe its C

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3 years ago

A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is

neonofarm [45]

Answer:

a) 25.5°(south of east)

b) 119 s

c) 238 m

Explanation:

solution:

we have river speed $v_{r}$ =2 m/s

velocity of motorboat relative to water is $v_{m/r}$ =4.2 m/s

so speed will be:

a) $v_{m}$ = $v_{r}$ + $v_{m/r}$

solving graphically

$v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}$

=4.7 m/s

Ф= $tan^{-1} (\frac{v_{r}}{v_{m/r}} )$

=25.5°(south of east)

b) time to cross the river: t= $\frac{w}{v_{m/r}}$ = $\frac{500}{4.2}$ =119 s

c) d= $v_{r}t$ =(2)(119)=238 m

note :

pic is attached

6 0

3 years ago