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3 years ago

Physics Conversion help!!

Physics

1 answer:

stiv31 [10]3 years ago

5 0

[Assuming that you've written 3.40 kg in 'a', and not 3.90 kg]

(a) 3,400 g x 0.001 = 3.40 kg [converting grams to kilograms]

(b) 220 cm x 0.01 = 2.2 m [converting centimeters to meters]

(d) 6.53 m x 100 = 653 cm [converting meters to centimeters]

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How long does it take to travel a distance of 672km at a speed of 95km/h?

Brilliant_brown [7]

Answer:

7.07 hours

Explanation:

divide the distance by the speed

so in this case, divide 672 by 95

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3 years ago

Your oven has a power rating of 5000 watt...

romanna [79]

A). 1,000 watts = 1 kilowatt
5,000 watts = 5 kilowatts

b). (5 kilowatts) x (2 hours) = 10 kilowatt-hours (kWh)

c). (10 kWh) x (15 cents/kWh) = $1.50

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3 years ago

Practice 3: Label the correct phase that would result from the Moon and Earth in these positions.

Anna71 [15]

Answer:

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3 years ago

Read 2 more answers

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

Burka [1]

Answer:

a) $v(2) = 3.9\,\frac{m}{s}$ , b) $v(4) = -15.7\,\frac{m}{s}$

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

$v(t) = 23.5 -9.8\cdot t$

The velocities at given instants are, respectivelly:

$v(2) = 3.9\,\frac{m}{s}$

$v(4) = -15.7\,\frac{m}{s}$

8 0

3 years ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0

3 years ago