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AnnyKZ [126]
2 years ago
15

How s the weather like​

Engineering
1 answer:
scoray [572]2 years ago
3 0

weather is like corono everywhere #stay safe

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Which tool ensures that a fastener has the proper amount of tightness
Sidana [21]

A torque wrench tool is a tool that ensures that a fastener has the proper amount of tightness.

<h3>What is the torque wrench used for?</h3>

The torque wrench tool is used to ensure screws and bolts are properly tightened. When performing home repairs and maintenance of equipment it is quite important that a torque wrench is used in other to prevent a scenario where a fastener (screws and bolts) does not become loose leading to equipment failure or damage. Because of its many advantages, this tool is often found in the possession of construction workers.

You can learn more about the benefits of a torque wrench tool here

brainly.com/question/15075481

#SPJ1

7 0
1 year ago
Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5s
zlopas [31]

Answer:

4m/s

Explanation:

We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted

Mathematically

Power_{motor} } =\frac{Energy }{time}\

We also know that Power = force x velocity      ..................(i)

The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load

=> power = mg x Velocity........(ii)

While hoisting the load at at constant speed only the potential energy of the mass increases

Thus Potential energy = Mass x g x H...................(iii)

where

g = accleration due to gravity (9.81m/s2)

H = Height to which the load is hoisted  

Equating equations (ii) and (iii) we get

m x g x v = \frac{mgh}{t}

thus we get v = H/t

Applying values we get

v = 6/1.5 = 4m/s

5 0
3 years ago
Free random points because im bored
tamaranim1 [39]

Answer:

thx for points B)

Explanation:

8 0
2 years ago
Read 2 more answers
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
viktelen [127]

Answer:

\theta_1=15^o\\\theta_2=75^o

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

V_x=V_{ox}=V_ocos\theta

Where \theta is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

V_y=V_{oy}-gt=V_osin\theta-gt

The  horizontal and vertical distances are, respectively:

x=V_{o}cos\theta t

\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

\displaystyle t_f=\frac{2V_osin\theta}{g}

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

\displaystyle R=\frac{V_o^2sin2\theta}{g}

Let's solve for \theta

\displaystyle sin2\theta=\frac{R.g}{V_o^2}

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}

\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}

Or equivalently:

\theta_2=90^o-\theta_1

Given Vo=37 m/s and R=70 m

\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}

\theta_1=15^o

And

\theta_2=90^o-15^o=75^o

5 0
3 years ago
A NC drill press is to perform a series of through-hole drilling operations on a 1.75 in thick aluminum plate that is a componen
jekas [21]

Answer:

26.7 min

Explanation:

First, we will find the <u>time required to drill each hole</u>:

  • N = 300 x 12/0.75 \pi = 1527.7 rev/min
  • fr = 1527.7 (0.015) = 22.916 in/min

Formula for <u>distance per hole</u>: 0.5 + A + 1.75

  • A = 0.5 (0.75) tan (90-100 / 2) = 0.315 in
  • Tm = (0.5 + 0.315 + 1.75) / 22.916 = 0.112 min

Now, we will calculate the <u>time required to draw back the drill form hole</u>:

              = 0.112 / 2 = 0.056 min

Time to move between holes = 1.5 / 15 = 0.1 min

For 100 holes, the number of moves between holes = 99

Total time required to drill 100 holes (t):

                       t = 100 (0.112 + 0.056) + 99 (0.1) = 26.7 min

7 0
3 years ago
Read 2 more answers
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