Ask question

Ask question

All categories

3 years ago

5

The initial velocity of a servo motor is 450 rpm, and the gear ratio of the gears in the motor is 2/5. What is the new velocity

of the motor if it has a
better torque?

1 answer:

aksik [14]3 years ago

7 0

Answer:

Depending on the amount of torque that is put in the motor, the new velocity is 550 rpm.

You might be interested in

An uncharged capacitor and a resistor are connected in series to a source of voltage. If the voltage = 7.41 Volts, C = 11.5 µFar

Musya8 [376]

Answer:

a) RC = 1.03 mseg.

b) Qmax = CV = 85.2 μC

c) Q = 53.9 μC

Explanation:

a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.

This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.

In this case , it can be calculated as follows:

ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.

b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:

Q = C V = 11.5 μF . 7.41 V = 85.2 μC

c) During the charging process, the charge increases following this equation:

Q = CV (1 - e⁻t/RC)

When t = RC, the expression for Q is as follows:

Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC

6 0

4 years ago

An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100

Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

$\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}$

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ( $R$ ) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$ , the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.

3 0

3 years ago

Do any of you have an email to contact brainly support? The contact us does not seem to work. Thanks!

Marysya12 [62]

Answer:

I think that if you go to the settings page, there should be a "contact us" option.

Hope this helped!! :)

3 0

3 years ago

50 for brainliest HELP ASAP<br> absurd answers will be recorded

s2008m [1.1K]

Answer:

1) This is because too much fuel is needed to get a payload from the surface to orbital altitude an accelerated to orbital speed.

2) This is because space travel present extreme environment that affect machines operations and survival.

Explanation:

Hope it helps

3 0

3 years ago

Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i

DedPeter [7]

Answer:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

Explanation:

The internal energy is equal to:

$dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV$

The entropy is equal to:

$dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV$

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

$\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }$

Replacing:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

4 0

3 years ago