Using the data from the table, what is P(3)?!

Tech A says that the brake pedal uses leverage to multiply foot pressure. Tech B says that when braking hard while moving

Nikolay [14]

Tech- A is correct

Explanation:

Leverage is defined as using a tool to gain mechanical influence. The measure of the benefit gained depends on what kind of lever is used and how it is utilized.
Leverage is designed in such a way that it can reproduce the force from your leg many times before any force is transferred to brake fluid.
The brake pedal size and the measure of leverage received depends on the overall design of the brake system.
The second-order lever is used in the brake pedal. The brake pedal applies leverage to populate the force employed to the master cylinder. The effort needed to drive a load depends on the corresponding distance of the load and the work from the fulcrum. The proportion of load and work is known as mechanical advantage.

7 0

3 years ago

The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re

Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 % , Irreversible

b) W_cycle = 600 KW , n_th = 100 % , Impossible

c) W_cycle = 400 KW , n_th = 66.67 % , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

TL = 400 K

TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

n_th < n_max ...... irreversible

n_th = n_max ...... reversible

n_th > n_max ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And, n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 400 = 200 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 200 / 600 = 33.33 %

- The type of process according to second Law of thermodynamics:

n_th = 33.333 % n_max = 66.67 %

n_th < n_max

Hence, Irreversible Process

b) QH = 600 kW, QC = 0 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 0 = 600 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 600 / 600 = 100 %

- The type of process according to second Law of thermodynamics:

n_th = 100 % n_max = 66.67 %

n_th > n_max

Hence, Impossible Process

c) QH = 600 kW, QC = 200 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 200 = 400 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 400 / 600 = 66.67 %

- The type of process according to second Law of thermodynamics:

n_th = 66.67 % n_max = 66.67 %

n_th = n_max

Hence, Reversible Process

7 0

3 years ago

How would you describe what would happen to methane if the primary bonds were to break?

erastova [34]

Answer:

All the bonds in methane (CH4CH4) are equivalent, and all have the same dissociation energy.

The product of the dissociation is methyl radical (CH3CH3). All the bonds in methyl radical are equivalent, and all have the same dissociation energy.

The product of that dissociation is methylene (CH2CH2). All the bonds in methylene are equivalent, and all have the same dissociation energy.

The product of that dissociation is methyne (CHCH) .

The C-H bonds in methane do not have the same dissociation energy as C-H bonds in methyl radical, which in turn do not have the same dissociation energy as the C-H bonds in methylene, which are again different from the C-H bond in methyne.

If (by some miracle) you were able to get all four bonds in methane to dissociate absolutely simultaneously, they would all show the same dissociation energy… but that energy, per bond broken, would be different than the energy required to break just one C-H bond in methane, because the products are different.

(In this case, it’s CH4→C+4HCH4→C+4H versus CH4→CH3+HCH4→CH3+H.)

To alter hydrocarbons you add enough energy to break a C-H bond. Why does only one bond break? What concentrates the energy on one C-H bond?

the weakest CH bond is the one that breaks. in plain alkanes it has to do with the molecular orbital interactions between neighboring carbon atoms. look at propane for example. the middle carbon has two C-C bonds, and each of those C-C bonds is strengthened by slight electron delocalization from the C-H bonds overlapping with the antibonding orbitals of the adjacent carbons.

since the C-H bonds on the middle carbon donate electron density to both of its neighbors, those two are weakest.

one of them will break preferentially.

which one actually breaks depends on the reaction conditions (kinetics). frankly it's whichever one ramdomly approaches a nucleophile first. when the nucleophile pulls of one of the H's, the other C-H bonds start to share (delocalize) the negative charge across the whole molecule. so while the middle C feels the majority of the negative charge character, the other two C's take on a fair amount as well...

by the way, alkanes don't really like to break and form anions like that.

a better example would be something like isopropyl iodide, where the C-I bond breaks and the I carries away the electron pair, forming a carbocation (also not particularly stable, but more so than the carbanion).

7 0

3 years ago

8. The operation of a TXV is controlled by the

Katena32 [7]

Answer should be C hopefully

4 0

3 years ago

A vertical piston-cylinder device initially contains 0.2 m3 of air at 20°C. The mass of the piston is such that it maintains a c

Ann [662]

Answer:

Amount of air left in the cylinder=m $_{2}$ =0.357 Kg