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3 years ago

What type of diagrams are used to present the conditions that are available to control outputs based on inputs?

Physics

2 answers:

NISA [10]3 years ago

8 0

Answer:

Explanation:

I did this. already. good luck!

creativ13 [48]3 years ago

4 0

Answer:

Explanation:

because it shows how this gets in

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Answer and explanation please!!

expeople1 [14]

Answer:

Option 3

Explanation:

O Option C is NEGATIVELY CHARGED, meaning it has GAINED ELECTRONS resulting in a GREATER number of ELECTRONS than PROTONS.

8 0

2 years ago

Read 2 more answers

An open pipe on an organ creates

PilotLPTM [1.2K]

Answer: 0.0163

Explanation: fn=n x v/2L

Fundamental frequency =f1

1 x (343/2L) = 10500

Rearrange the equation

L= v/(2xFn)

L= 343/(2x10500)

L=0.0163

8 0

2 years ago

The y-component of the force F which a person exerts on the handle of the box wrench is known to be 70 lb. Determine the x-compo

Yuri [45]

Answer:

Fx = 121.24lb

F = 140lb

Explanation:

Since we are not given the angles subtended by the force, we can assume it to be 30 degrees.

The y component of the force expressed by the formula:

Fy = Fsintheta

Given the y-component of the force F to bee 70lb

70lb = Fsintheta

Get magnitude of the force

F = 70/sin theta

F = 70/sin 30

F = 70/0.5

F = 140lb

Get the x-component of the force

Fx = Fcos theta

Fx = 140cos 30

Fx = 140(0.8660)

Fx = 1,212.4lb

Hence the x-component of the force sis 121.24lb

Note that the angle used was assumed. Other values can as well be used

5 0

2 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.