Phil wants to make a dark themed superhero movie. What could be his target demographic?

How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.

Snowcat [4.5K]

Answer:

$N_3=\dfrac{T_1}{T_3}N_1$

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

Lets take

$Speed\ of\ gear 1=N_1$

$Number\ of\ teeth\ of\ gear 1=T_1$

$Speed\ of\ gear 3=N_3$

$Number\ of\ teeth\ of\ gear 3=T_3$

All external matting gears will rotates in opposite direction with respect to each other.

So the speed of gear third can be given as follows

$\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}$

$N_3=\dfrac{T_1}{T_3}N_1$

3 0

2 years ago

A(n)_____ is a device that provides the power and motion to manipulate the moving parts of a valve or damper used to control flu

Lesechka [4]

Answer:

Out of the four options provided

option A. actuator

is correct

Explanation:

An actuator is the only device out of the four mentioned devices that provides power and ensures the motion in it in order to manipulate the movement of the moving parts of the damper or a valve used whereas others like ratio regulator are used to regulate air or gas ratio and none mof the 3 remaining options serves the purpose

5 0

2 years ago

The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The

klemol [59]

The power that must be supplied to the motor is 136 hp

<u>Explanation:</u>

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power = ?

According to the equation of motion:

F = ma

$3(500) - 1000 = \frac{1000}{32.2} * a$

a = 16.1 ft/s²

We know,

$v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s$

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

= 136 hp

Therefore, the power that must be supplied to the motor is 136 hp

5 0

3 years ago

What components are included in a basic engine block?

atroni [7]

Houses the CYLINDERS, Water Jacket & Crankcase

8 0

2 years ago

P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si

klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V 0.184 W

2.499 mV 125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

$V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\$ *5

= 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:

V_o = A_voc*V_i*(R_o/R_l+R_o)

= 4.545*(100/100+50)

= 3.03 V

Now we can determine delivered power:

P_L = V_o^2/R_L

= 0.184 W

Apply voltage-divider principle to the circuit:

V_o = (R_o/R_o+R_s)*V_s

= 50/50+100*10^3*5

= 2.499 mV

Now we can determine delivered power:

P_l = V_o^2/R_l

= 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.

4 0

2 years ago