What is the tool used to take off a wheel nut

Used ______ must be hot drained for 12 hours or crushed before disposal.

WITCHER [35]

Answer:

A

Explanation:

The answer is towels because towels after a little bit of just sitting around have a chemical reaction that cam cause them to spontaneously combust

5 0

3 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$

We can assume the same drag coeeficient $C_{D1}=C_{D2}=C_{D}$ and we got:

$\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D} \frac{\rho}{2}V^2 (wh) V$

$\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})$

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

$R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}$

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

$\frac{l}{h}=\frac{5}{1.7}2.941$

And we can estimate the calue of $C_D = 1.2$ from a figure.

And we can calculate the power difference like this:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

8 0

3 years ago

A current I flows in the inner conductor of an infinitely long coaxial line and returns via the outer conductor. The radius of t

scZoUnD [109]

Answer:

See explaination

Explanation:

By definition, we can say that Magnetic flux density is defined as the amount of magnetic flux in an area taken perpendicular to the magnetic flux's direction. An example of magnetic flux density is a measurement taken in teslas.

Please kindly check attachment for the step by step solution of the given problem.

8 0

3 years ago

Suppose we would like to predict the duration of an eruption of the Old Faithful geyser in Yellowstone National Park based on th

liq [111]

Answer:

head(faithful)

faithful_model<-lm(eruptions~waiting,data=faithful)

summary(faithful_model)

Explanation:

6 0

3 years ago

El tiempo hasta que falle un sistema informático sigue una distribución Exponencial con media de 600hs. (Utilice 3 decimales par

Lesechka [4]

Answer:

La probabilidad pedida es $0.717$

Explanation:

Primero comencemos definiendo la variable aleatoria. Para nuestro problema, la variable aleatoria es la siguiente :

$X:$ '' El tiempo (en horas) hasta que falle un sistema informático ''

La variable aleatoria $X$ será entonces una variable aleatoria continua.

Sabemos que sigue una distribución exponencial con una media de 600 hs.

Esto se escribe :

$X$ ~ ε ( λ ) (I)

En donde λ es igual a la inversa de la media. Esto se escribe :

λ $=\frac{1}{E(X)}$

En donde $E(X)$ es la media de la variable. Por ende, si reemplazamos los datos del ejercicio obtenemos ⇒

λ $=\frac{1}{E(X)}$ ⇒ λ $=\frac{1}{600}$

Si reemplazamos el valor de λ en (I) obtenemos :

$X$ ~ ε $(\frac{1}{600})$

La función de distribución de $X$ (por ser una variable aleatoria exponencial) es :

$F_{X}(x)=P(X\leq x)=$ 1 - e ^ ( - λx) con x > 0 y $F_{X}(x)=0$ en caso contrario.

Si reemplazamos el valor de λ en la función de distribución de $X$ obtenemos :

$F_{X}(x)=P(X\leq x)=1-e^{-\frac{x}{600}}$

Dado que la variable aleatoria $X$ se distribuye de manera exponencial, el hecho de saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos aporta ninguna información sobre lo que ocurrirá después. Esta característica se conoce como propiedad de perdida de memoria de la variable aleatoria exponencial. Entonces, la probabilidad pedida se reduce a calcular :

$P(X>200)$

Dado que saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos dice nada sobre lo que ocurrirá instantes posteriores a esas 400 hs.

Calculamos entonces la probabilidad pedida :

$P(X>200)=1-P(X\leq 200)=1-F_{X}(200)=1-(1-e^{-\frac{200}{600}})=e^{-\frac{1}{3}}=0.717$

7 0

3 years ago