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3 years ago

How long does it take to complete Fort nite: Save The World?

Engineering

2 answers:

yawa3891 [41]3 years ago

6 0

It takes long so a whole week if you stay focused

maria [59]3 years ago

4 0

<h3><u>Answer:</u></h3>

Over 8 straight days of playing

<h3><u>Explanation:</u></h3>

200+ hours, if you're a game master

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List and describe three basic Boolean operations commonly used in

valentinak56 [21]

Answer:

There are three basic Boolean operations: JOIN (Union), CUT (Difference), and INTERSECT. ... The CUT operation subtracts the volume of one solid object from the other solid object. The INTERSECT operation keeps only the volume common to both solid objects.

4 0

3 years ago

Which of the following statements about static electricity is true?

andreev551 [17]

D,B I thinks do check✨

6 0

3 years ago

The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra

Lina20 [59]

Answer:32.4m/ $s^2$

Explanation:

Given data

$radius\left ( r\right )$ =0.4m

Intial angular velocity $\left ( \omega_0\right )$ =4rad/s

angular acceleration $\left ( \alpha\right )$ =5rad/ $s^2$

angular velocity after 1 sec

$\omega$ = $\omega_0$ + $\alpha\times\t$

$\omega$ =4+5 $\left ( 1\right )$

$\omega$ =9rad/s

Velocity of point on the outer surface of disc $\left ( v\right )$ = $\omega_0\timesr$

v= $9\times0.4$ m/s=3.6m/s

Normal component of acceleration $\left ( a_c\right )$ = $\frac{v^2}{r}$

$a_c$ = $\frac{3.6\times3.6}{0.4}$ =32.4m/ $s^2$

3 0

4 years ago

What does the DHCP server configures for each host?

lidiya [134]

Answer: IP address

Explanation:

DHCP server automatically assigns an IP address and other information to each host on the network so they can communicate efficiently with other endpoints

8 0

4 years ago

Find an expectation value in the n th state of the harmonic oscillator.Find an expectation value in the n th state of the harmon

s344n2d4d5 [400]

The classical motion for an oscillator that starts from rest at location x₀ is

x(t) = x₀ cos(ωt)

The probability that the particle is at a particular x at a particular time t

is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average

to get the spatial density. Our natural time scale for the averaging is a half

cycle, take t = 0 → π/ ω

Thus,

ρ = $\frac{1}{\pi / w} \int\limits^\pi_0 {d(x - x_o cos(wt))} \, dt$

Limit is 0 to π/ω

We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt) so that

ρ(x) = $-\frac{w}{\pi } \int\limits^x_x {\frac{d ( x - y)}{x_ow sin(wt)} } \, dy$

Limit is x₀ to -x₀

$\frac{1}{\pi } \int\limits^x_x {\frac{d (x-y)}{x_o\sqrt{1 - cos^2(wt)} } } \, dy$

Limit is -x₀ to x₀

$= \frac{1}{\pi } \int\limits^x_x {\frac{d(x-y)}{\sqrt{x_o^2 - y^2} } } \, dy\\ \\= \frac{1}{\pi\sqrt{x_o^2 - x^2} }$

This has $\int\limits^x_x {p(x)} \, dx = 1$ as expected. Here the limit is -x₀ to x₀

The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.

6 0

4 years ago