By balancing information security and access, a completely secure information system can be created.A. TrueB. False

Would you ever date a transgender person??

nekit [7.7K]

Answer:

well my bff is trans so yes I guess xd

4 0

3 years ago

Read 2 more answers

A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of

lara [203]

Answer:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

$e = 1 -frac{T_C}{T_H}$

We have on this case after convert the temperatures in kelvin this:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

And the maximum power output on this case would be defined as:

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Where $Q_H$ represent the heat associated to the deposit with higher temperature.

4 0

3 years ago

A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts

lions [1.4K]

Answer:

The coefficient of static friction, μₛ, between the trunk and turntable = 0.32

Explanation:

For this motion of the trunk B,

Initial velocity, v₀ = 0

Tangential Acceleration, a = 0.28 m/s²

Time taken, t = 10s

Using equations of motion,

v = v₀ + at

v = 0 + 0.28 × 10 = 2.8 m/s

Frictional force, Fᵣ = μₛN

μₛ = coefficient of static friction,

N = Normal reaction exerted on the trunk B as a result of its weight = mg

Doing a force balance on the trunk B,

Force keeping the trunk B in circular motion must balance the frictional forces.

Force keeping the trunk B in circular motion, F = mv²/r

Fᵣ = F

μₛN = mv²/r but N = mg

μₛmg = mv²/r

μₛg = v²/r

μₛ = v²/gr

μₛ = 2.8²/(9.8 × 2.5) = 0.32

Hope this helps!!!

3 0

3 years ago

An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is

baherus [9]

Answer:

the correct distance is 202 ft

Explanation:

The computation of the correct distance is shown below:

But before that correction to be applied should be determined

= (101 ft - 100 ft) ÷ (100 ft) × 200 ft

= 2 ft

Now the correct distance is

= 200 ft + 2 ft

= 202 ft

Hence, the correct distance is 202 ft

The same would be relevant and considered too

4 0

3 years ago

Determine the combined moment about O due to the weight of the mailbox and the cross member AB. The mailbox weighs 3.2 lb and th

koban [17]

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

The perpendicular distance from Y axis of centroid of the mailbox

= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

= 52.8 + 133.9 = 186.7 Ib-in

8 0

3 years ago