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2 years ago

How many newtons does a 5-kg backpack weigh on Earth?

Physics

1 answer:

zvonat [6]2 years ago

7 0

A 5kg backpack will weigh 49 newtons on earth

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If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add

mr Goodwill [35]

Each $Al^+^3$ ion contains three extra protons. Hence, the extra charge on each $Al^+^3$ = $3 \times 1.6 \times 10^-^1^9$ C

Total charge = 0.035 pC

Total charge (Q) = $0.035 \times 10^-^1^2$ C

Let the number of $Al^+^3$ ions be n.

According to question:

$n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2$

$n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}$

$n = 7.29167 \times 10^4$

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0

2 years ago

Would it be m/s or kg?

Nesterboy [21]

Answer:

m.s

Explanation:

3 0

2 years ago

At the accounting break-even point, Swiss Mountain Gear sells 22,940 ski masks at a price of $19 each. At this level of producti

gogolik [260]

Answer:

fixed cost $231220

Explanation:

given data:

quantity of skl mask = 22,940

skl mass rate = $19 each

variable cost = $6 per unit

therefore

Contribution/unit of ski mask = 19 - 6 = $13

for 22940 ski masks, total cost = (22940*13) = 298220

Depreciation cost $67000

fixed cost is = 298220 - 67000 = $231220

8 0

2 years ago

A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a

andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

$sin\theta = 1.22\frac{\lambda}{d}$

Here,

$\lambda$ = Wavelength

d= Diameter of aperture

$\theta$ = Angular resolution or diffraction angle

Our values are given as,

$\theta = 11\°$

The frequency of the sound is $f = 9100 Hz$

The speed of the sound is $v = 343 m/s$

The wavelength of the sound is

$\lambda = \frac{v}{f}$

Here,

v = Velocity of the wave

f = Frequency

Replacing,

$\lambda = \frac{(343 m/s)}{(9100 Hz)}$

$\lambda = 0.0377 m$

The diffraction condition is then,

$sin\theta = 1.22\frac{\lambda}{d}$

Replacing,

$sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}$

d = 0.24 m

Therefore the diameter should be 0.24m

6 0

2 years ago

If you could be the first person to go somewhere where would it be and why??? Example plz

vivado [14]

New York because of the lady of all the people and excitement

3 0

2 years ago

Read 2 more answers