What is the acceleration of a 50 kg object pushed with a net force of 500 newtons?

Flying against the wind, an airplane travels 2670 km in 3 hours. Flying with the wind, the same plane travels 11,070 km in 9 hou

xxMikexx [17]

Answer:

speed of plane in still air = 1060 km/h

speed of wind = 170 km/h

Explanation:

Let teh speed of plane in still air is vp and the speed of air is va.

Irt travels 2670 km in 3 hours against the wind

So,

vp - va = 2670 / 3 = 890 km/h ..... (1)

It travels 11070 km in 9 hours along the wind.

vp + va = 11070 / 9 = 1230 km/h .... (2)

Adding both the equations

2 vp = 2120

vp = 1060 km/h

and va = 1230 - vp = 1230 - 1060 = 170 km/h

5 0

3 years ago

A heat engine operates between two reservoirs at 300K. During each cycle, it absorbs 200 calories from the high temperature rese

yanalaym [24]

(a) Zero

The maximum efficiency (Carnot efficiency) of a heat engine is given by

$\eta=1-\frac{T_C}{T_H}$

where

$T_C$ is the low-temperature reservoir

$T_H$ is the high-temperature reservoir

For the heat engine in the problem, we have:

$T_C = 300K$

$T_H = 300K$

Therefore, the maximum efficiency is

$\eta=1-\frac{300}{300}=0$

(b) Zero

The efficiency of a heat engine can also be rewritten as

$\eta = \frac{W}{Q_H}$

where

W is the work performed by the engine

$Q_H$ is the heat absorbed from the high-temperature reservoir

In this problem, we know

$\eta=0$

Therefore, since the term $Q_H$ cannot be equal to infinity, the numerator of the fraction must be zero as well, which means

W = 0

So the engine cannot perform any work.

5 0

2 years ago

Suppose you need your silicon circuit element to run continuously for 3 minutes before it shuts off long enough to cool back dow

Zigmanuir [339]

The maximum rate at which energy can be added to the circuit element mathematically given as

$MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}$

<h3>What is the maximum rate at which energy can be added to the circuit element?</h3>

Generally, the equation for P is mathematically given as

$P=\ln s \frac{\Delta T}{\Delta t}$

Therefore

$Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}$

$\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}$

$\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}$

Max temp Change

$MaxT=5.6^{\circ} \mathrm{C}$

$\text { time }=3 \times 60$

t=180s

In conclusion, Max Energy Rate

$MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}$

$MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}$