Ask question

Ask question

All categories

3 years ago

8

Quinn’s relatives relayed a story about putting on a headset and seeing a digital world that they could walk around in and explo

re in order to see what ancient Benin looked like, as if they were really there. What is this called?
systems software
hardware
virtual reality
augmented reality

2 answers:

Kryger [21]3 years ago

6 0

Answer:

I know it is C)Virtual reality

Explanation:

Look at the clues

story about putting on a headset ( virtual reality head set!)

seeing a digital world (A virtual reality world)

they could walk around in (Fake walking you are basically jogging in place)

explore in order to see what ancient Benin looked like (Looking at a real place only digitally)

as if they were really there ( they think they are actually there)

The only reason I know all of this is because I have done virtual reality multiple times and I LOVED it SUPER fun ( I was doing archery) :) Hope this helps!

Ket [755]3 years ago

4 0

Answer:

C)Virtual reality

Explanation:

You might be interested in

the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the i

Serggg [28]

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

$V(t) = 5(1-e^{-0.5t})$

The current flowing through the inductor is given by

$i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)$

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

$i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\$

$i(t) = 1000t +2000e^{-0.5t} -2000\\$

So the current at t = 2 seconds is

$i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A$

The energy stored in the inductor at t = 2 seconds is

$W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J$

4 0

3 years ago

Visual aids are useful for all of the following reasons except

Ksivusya [100]

Answer:

B

Explanation:

their presence allows the speaker to take a rest from talking

I couldnt see the question

4 0

2 years ago

A long corridor has a single light bulb and two doors with light switch at each door.

Zigmanuir [339]

Answer:

The answer is below

Explanation:

Let A represent the first switch, B represent the second switch and C represent the bulb. Also, let 0 mean turned off and 1 mean turned on. Since when both switches are in the same position, the light is off. This can be represented by the following truth table:

A B C (output)

0 0 0

0 1 1

1 0 1

1 1 0

The logic circuit can be represented by:

C = A'B + AB'

The output (bulb) is on if the switches are at different positions; if the switches are at the same position, the output (bulb) is off. This is an XOR gate. The gate is represented in the diagram attached below.

6 0

3 years ago

The following median grain size data were obtained during isothermal liquid phase sintering of an 82W-8Mo-8Ni-2Fe alloy. What is

Morgarella [4.7K]

Answer:

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

( $d^{2}$ - $d_{0} ^{2}$ = kt )

Explanation:

The plot attached below shows the time dependence of the growth of grain.

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

the ideal growth follows this principle = $d^{2} - d^{2} _{0}$ = kt

d = final grain size

$d_{0}$ = initial grain size

k = constant ( temperature dependent )

t = 0

8 0

3 years ago

An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t

solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the $w_{p}$ to h₁, where $w_{p}$ is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute $w_{p}$ To computer

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water $v_{f}$ = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ = 0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

= 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

= 417.5 + 0.001043 (14900)

= 417.5 + 15.5407

= 433.04 kJ/kg

Find h₃

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = $s_{f}$ + $x_{4} s_{fg}$

$x_{4} = s_{4} -s_{f} /s_{fg}$

$x_{4}$ = 6.14 - 2.45 / 3.89

$x_{4}$ = 0.9497

The enthalpy h₄:

h₄ = $h_{f} +x_{4} h_{fg}$

= 908.4 + 0.9497(1889.8)

= 908.4 + 1794.7430

= 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅ = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = $s_{f}$ + $x_{6} s_{fg}$

$x_{6} = s_{6} -s_{f} /s_{fg}$

$x_{6}$ = 7.29 - 1.3028 / 6.0562

$x_{6}$ = 0.988

The enthalpy h₆:

h₆ = $h_{f} +x_{6} h_{fg}$

= 417.51 + 0.988 (2257.5)

= 417.51 + 2230.41

h₆ = 2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

$P_{p}$ is found by using:

mass flow rate = m = 1.74 kg/s

Volume of water = v₁ = 0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

$P_{p}$ = ( m ) ( v₁ ) ( p₂ - p₁ )

= (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

= (1.74 kg/s) (0.001043 m³/kg) (14900)

= 27.04

$P_{p}$ = 27 kW

Compute heat added $q_{a}$ and heat rejected $q_{r}$ from boiler using computed enthalpies:

$q_{a}$ = ( h₃ - h₂ ) + ( h₅ - h₄ )

= ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

= 2726 + 655

= 3381 kJ/kg

$q_{r}$ = h₆ - h₁

= 2648 kJ/kg - 417.5 kJ/kg

= 2232 kJ/kg

Compute net work

W $_{net}$ = $q_{a}$ - $q_{r}$

= 3381 kJ/kg - 2232 kJ/kg

= 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m = 1.74 kg/s

W $_{net}$ = 1150 kJ/kg

P = m * W $_{net}$

= 1.74 kg/s * 1150 kJ/kg

= 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

= 1.74 kg/s * 655

= 1140 kW

Compute Thermal efficiency of this system

μ $_{t}$ = 1 - $q_{r}$ / $q_{a}$

= 1 - 2232 kJ/kg / 3381 kJ/kg

= 1 - 0.6601

= 0.34

= 34%

7 0

3 years ago