Which metal has the ability to rust

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

6 0

3 years ago

Determine the minimum number of 120-volt, 20-ampere circuit breakers for a continuous load consisting of 63 feet of track lighti

maks197457 [2]

If we have 20-ampere circuit breakers. The number of circuits to the larger whole number is: 13.

<h3>Number of circuits</h3>

Receptacles on a single strap= 180 VA each.

Hence,

VA of the circuit=(Volts x Amperes)/One receptacle

Let plug in the formula

VA of the circuit=(120 volts x 20 amperes)/180 VA

VA of the circuit= 2,400 VA (circuit)/180 VA

VA of the circuit = 13 circuits

Therefore the number of circuits to the larger whole number is: 13.

Learn more about number of circuits here:brainly.com/question/2969220

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4 0

2 years ago

Which of the following can minimize engine effort in save fuel

pentagon [3]

Answer:

hmm

Explanation:

How to Save Fuel

Keep your vehicle's engine in good condition.

Maintain proper air pressure in your tires.

The faster you travel, the greater your fuel consumption is.

Try to speed up gradually when you stop for a traffic light.

Try to drive smoothly.

Do not allow your engine to idle.

hope this helps

7 0

3 years ago

Read 2 more answers

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

3 years ago

a(n) ? is an intentionally constructed, low-impedance electrically conductive path designed and intended to carry current during

tester [92]

A effective ground-fault current path is an intentionally constructed, low-impedance electrically conductive path designed and intended to carry current during ground-fault conditions from the point of grounding on a wiring system to the electrical supply source.

<h3>Is earth an effective ground fault current path?</h3>

Sticking the wire in the ground is not sufficient since the earth is not thought to be a reliable ground-fault current channel.
The electrical system of a building or other structure is based on grounding.
To give a fault current a secure path to travel, grounding is used.
When installing switches, light fixtures, appliances, and receptacles, a complete ground route must be kept.
The undesired current flow trips circuit breakers or blows fuses in a system that is correctly grounded.
Through the use of a grounding bank, effective grounding maintains voltages within predetermined limits during a line-to-ground fault (short-circuit condition).

To learn more about ground-fault current channel refer,

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5 0

1 year ago