Let's say the distance is D. Then the time going is D/10 sec. The time returning is D/20 s. The total time is 3D/20 s, and the total distance is 2D. The average speed for the round trip is (total distance)/(total time). That's (2D) ÷ (3D/20). That's (40D/3D) which is 13-1/3 m/s. (I thought it was going to depend on the distance, but it doesn't.)

5 0

3 years ago

Benjamin recently issued new mobile phones to the marketing team at his company. Each phone can transmit encrypted information f

Misha Larkins [42]

Complete Question:

Which technology is set up on the marketing team's phones?

a) Bluetooth

b) Tethering

c) NFC

d) Hotspot

Answer:

NFC

Explanation:

Near Field Communication (NFC) is a short-range wireless connectivity standard that uses magnetic field induction to enable communication between devices when brought within a few centimetres from each other. NFC offers a secured communication between connected electronic devices with simple setup that can be used to bootstrap more-capable wireless connections. It only covers a range of 10 cm to 20 cm.

None of the other options can transmit or share information for payments among the team except the NFC. NFC is also suitable for this setup because only a small range is expected to be covered.

4 0

3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago

What step precedes the onset of hydrogen fusion in a star?

Andrew [12]

If it is about the hydrogen fusion in main sequence stars, where the temperature of the center or the core rises the same time as the mass does, which results of the PP process is domination the process particularly in low masses. This is just one example of the step that percedes the onset of hydrogen fusion.

4 0

3 years ago