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3 years ago

8. In a closed hydraulic brake system, the hydraulic pressure:

Engineering

2 answers:

beks73 [17]3 years ago

8 0

Answer:

Option A is correct.

The hydraulic pressure in a closed hydraulic brake system is constant.

Explanation:

The hydraulic brake system works on the principle of pressure transmission. It is termed a closed hydraulic brake system when there is no leakage of fluid all through the setup. The transmission works on the basis of Pascal's law for pressure and it explains that pressure applied to a fluid is transmitted undiminished to every part of the fluid.

The force applied on the brake pedals by the driver at the controlling end of the hydraulic system is transmitted by the pushrod on the piston(s) in the master cylinder.

This causes fluid from the brake fluid reservoir to flow into a pressure chamber. This in turn causes an increase in the pressure of the entire hydraulic system, forcing fluid through the hydraulic lines toward the caliper(s) where it then acts upon the caliper piston(s).

The force applied on the brakes pedal, F₁,

is applied on the cross sectional Area of the master piston, A₁ and this is transmitted in terms of pressure by the fluid (with no loss in pressure, that is, constant pressure, if the hydraulic braking system is especially a closed one) through the callipers to the calliper(s) pistons.

Hope this Helps!!!

DiKsa [7]3 years ago

3 0

Answer:

A. Is constant

Explanation:

This is in accordance with pascal's law of pressure that states the hydraulic pressure in a fluid is transmitted undiminished to every portion of the fluid.

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3. Suppose you work for this company that evaluates Boolean circuits in exponential time. Since you are very smart, your manager

hammer [34]

Answer:

Ergr5

Explanation:

3 0

3 years ago

g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0

Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵ * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0

2 years ago

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in $2.25 * 10^4 BTU$ . It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - $1 BTU = 778.17 lbf*ft$

$2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft$

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:

$Ep = m*g*(h_2 - h_1)\\ W = m*g$