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3 years ago

14

Ok I need a new laptop but I'm not sure which one. The Surface laptop 3 is better in many fields like the camera, speaker, audio

, overall performance or could get the Macbook air which connects to apple tv has apple exclusive things like airdrop which is good since I already have couple of Apple products. I'm so confused and would love some opinions

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

get the surface laptop 3

Explanation:

its better quality and performs excellent!

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Which emission is created by the heat in the combustion process?

Musya8 [376]

Water vapor and carbon dioxide!

5 0

3 years ago

Read 2 more answers

a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem

Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

$C = 100nF = 100*10^-^9F$

a)Using the formula:

$F_c = \frac{1}{2pie*Rc}$

$8000= \frac{1}{2*3.14*R*100*10^-^9}$

$R =\frac{1}{2*3.14*100*10^-^9*8000}$

R = 199.04 ohms

b) diagram is attached

c) $H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}$

$H(F) = \frac{1}{1-j\frac{fc}{f}}$

At F = 20KHz and Fc= 8KHz we have:

$H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}$

$|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}$

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

5 0

3 years ago

Are ocean currents always cold

swat32

Answer:

The surface ocean currents have a strong effect on Earth's climate. ... However, these areas do not constantly get warmer and warmer, because the ocean currents and winds transport the heat from the lower latitudes near the equator to higher latitudes near the poles.

8 0

3 years ago

A 46.0-g meter stick is balanced at its midpoint (50.0 cm, zero point is a left end of stick). Then a 210.0-g weight is hung wit

Anna71 [15]

Clockwise torque due to 100g is 0.1029 Nm and 200g is 1.4406 Nm. Clockwise torque due to stick mass is 0.2254 Nm and Counter-clockwise torque due to normal force is 1.7689 Nm.

<h3>What is clockwise torque?</h3>

The right-hand rule for cross products determines the direction of torque, which is calculated as the cross product of force and distance. Your thumb will point in the direction of the torque if you place your palm in the direction of the applied force and extend your fingers from the pivot point in that direction.

A related right-hand rule relates the direction of the rotation to the direction of the torque. Your fingers will curl in the direction of rotation if you point your thumb in the direction of the torque.

Positive torques cause counter clockwise rotation, while negative torques cause clockwise rotation.

The sum of all torques must be zero at equilibrium since an object in equilibrium has no net torque.

When the force is applied in a direction perpendicular to the line connecting the pivot and the force, the torque is at its greatest.

You can calculate the torque's magnitude using

$\begin{displaymath}\tau =rF_{\bot }=rF\sin \theta .\end{displaymath}$

To solve problems involving torques, follow these eight steps: read the issue, create a free-body diagram, locate the pivot point, write down the expressions for all torques, For equilibrium conditions, set the sum of torques to zero, list all known variables, pick the desired variable(s), write down equations involving those variable(s), solve the equations, plug in numbers, and test your solution.

Clockwise torque due to 100 g ⇒ T1 = 0.105* 9.8* 0.1 = 0.1029 Nm

Clockwise torque due to 200 g ⇒ T2 = 0.210* 9.8* 0.7 = 1.4406 Nm

Clockwise torque due to stick mass ⇒ T3 = 0.046* 0.5* 9.8 =0.2254 Nm

Counter-clockwise torque due to normal force ⇒ T4 = (0.046 + 0.21 + 0.105)*9.8* 0.5 = 1.7689 Nm

Learn more about torque

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7 0

1 year ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago