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2 years ago

Swing arm restraints are intended to prevent a vehicle from falling off a lift.

Engineering

1 answer:

castortr0y [4]2 years ago

4 0

The question is asking whether that statement is true or false. Options are;

A) True

B) False

This is about usage of Swing arm restraints.

B) False

There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.

Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.

This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.

Read more at; brainly.com/question/17972874

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Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribu

aksik [14]

Answer:

Para x=0:

$\phi=1.2 W/m^{2}$

Para x=30 cm:

$\phi=-2.4 W/m^{2}$

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

$\phi=-k\frac{dT}{dx}$ (1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

$\frac{dT(x)}{dx}=300x-30$

Remplezando esta derivada en (1):

$\phi=-0.04(300x-30)$

Para x=0:

$\phi=0.04(30)$

$\phi=1.2 W/m^{2}$

Para x=30 cm:

$\phi=-0.04(300*0.3-30)$

$\phi=-2.4 W/m^{2}$

Espero que te haya ayudado!

4 0

3 years ago

If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det

Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

solution:

to find the shear stress

T_allow=T_fail/F.S

=120 Mpa/2.5

=48 Mpa

we know that,

V=P

Area for shear head:

A(head)=π×d×t

=π×0.04×0.075

=0.003×πm^2

Area for plate:

A(plate)=π×d×t

=π×0.08×0.03

=0.0024×πm^2

now we have to find shear stress for both head and plate

For head:

T_allow=V/A(head)

48 Mpa=P/0.003×π ..(V=P)

P =48 Mpa×0.003×π

=452.16 KN

For plate:

T_allow=V/A(plate)

48 Mpa=P/0.0024×π ..(V=P)

P =48 Mpa×0.0024×π

=361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

P=361.91 KN

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