3 years ago

Drag the tiles to the correct boxes to complete the pairs.

Engineering

1 answer:

Jobisdone [24]3 years ago

3 0

Answer:

Check image, right on PLATO

You might be interested in

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon

balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

$torque =\frac{power}{2\pi N}$

$power=2\pi N\times T$

P= $2\times \pi \times2500 \times 4.5$

P=70,685W

P=70.685 KW

power= $\frac{work done}{time}$

work done = power * time

= 70.685*30=2120.55J

= 2.12 kJ

7 0

3 years ago

EJERCICIO 6

Ket [755]

Answer:

mnfokfnfi3or

Explanation:

can you translate it into english.....

8 0

3 years ago

Suppose you want to find the sum of two sinusoidal voltages, given as follows: v1(t)=V1 cos(ωt+ϕ1) and v2(t)=V2 cos(ωt+ϕ2)v1(t)=

seropon [69]

Answer:

Attached is the full solution.

3 0

3 years ago

Read 2 more answers

As a parent, you have a responsibility to ensure the safety of your family. Which of the safety concerns listed below would be c

Furkat [3]

Answer:

B and C

Explanation:

4 0

3 years ago

A car is traveling at 70 mi/h on a level section of road with good, wet pavement. Its antilock braking system (ABS) only starts

siniylev [52]

Answer: The theoretical stopping distance is 1.245 miles (6572.193 ft).

Explanation:

First, the physical model is created by using Principle of Energy Conservation and Work-Energy Theorem. It is assumed that surface is horizontal, so there are no changes associated with potential energy. The car has an initial kinetic energy, which is completely dissipated by braking.

$K_{1} = \Delta W_{loss}$

$\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (\mu_{2} \cdot \Delta s_{2}+\mu_{2'} \cdot \Delta s_{2'})$

The distance is isolated from previous equation:

$\Delta s_{2'} = \frac{1}{\mu_{2'}}\cdot (\frac{1}{2}\cdot \frac{v^{2}}{g} - \mu_{2} \cdot \Delta s_{2})$

By replacing variables, the distance is calculated herein:

$\Delta s_{2'} = \frac{1}{\frac{0.02}{0.80}}\cdot (\frac{1}{2}\cdot \frac{(102.667\frac{ft}{s} )^2}{32.174 \frac{ft}{s^{2}}} - (0.02) \cdot (100 ft))$

$\Delta s_{2'} = 6472.193 ft\\\Delta s_{2'} = 1.226 mi$

The theoretical stopping distance is:

$\Delta s = \Delta s_{2} + \Delta s_{2'}\\\Delta s = 6572.193 ft\\\Delta s = 1.245 mi$

5 0

3 years ago