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3 years ago

What measurements or observations tells you that a car is accelerating

2 answers:

8 0

You need to observe the car at two different times.

-- The first time:
You write down the car's speed, and the direction it's pointing.

-- The second time:
You write down the car's speed and the direction it's pointing, again.

You take the data back to your lab to analyze it.

-- You compare the first and second speed. If they're different,
then the car had acceleration during the time between the two
observations.

-- You compare the first and second direction. If those are different,
even if the speeds are the same, then the car had acceleration during
the time between the two observations.

(Remember, "acceleration" doesn't mean "speeding up".
It means any change in speed or direction of motion.)

PtichkaEL [24]3 years ago

7 0

Frame of reference. It uses the second law of motion to determine whether it is moving or not...

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If r1 < r2 < r3, and if these resistors are connected in series in a circuit, which one dissipates the most power

Vesnalui [34]

All three have the same current, so that is not a factor. Wattage (power) is E*I or i^2 R. The higher the resistance, the more power dissipated. The answer is R3 because it has the highest resistance.
R3 <<<< ===== answer.

6 0

2 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago

Answer part (d) please

finlep [7]

Answer:

MARK me brainliest please and follow my page

Explanation:

All you have to do to get the average speed is to calculate the total distance covered and divide it by the total time taken

= 16/18 = 0.88m/s

3 0

2 years ago

Read 2 more answers

The emissivity of galvanized steel sheet, a common roofing material, is 0.13 at temperatures around 300 K, while its absorptivit

vfiekz [6]

Answer:

Total heat transfer rate is combined convection and radiation transfer rate, multiplied with absorptivity factor. From that equation you can find temperature of the roof using iteration method.

Explanation:

Known:

steel sheet roof emissivity ∈ = 0.13

steel sheet roof absorptivity $\alpha$ _s = 0.65

air temperature T∞ = 16°C ...= 289K

heat transfer rate $G_s$ = 750 W/m2

natural convection h = 7W/m2K

Total heat transfer rate is combined heat transfer rate of convection and radiation, multiplied with absorptivity factor:

$\alpha$ $G_s$ = $q_{conv} +q_{rad} =hA($ T∞- $T_m$ )+∈σA( $T_{surf} -T_{s} )$

$\alpha$ $G_s$ - $h(T_s-$ T∞)-∈σ( $T_{s} -T_{surf} )$ =0

2562-7 $T_s$ -7.371*10^-9=0

Use iteration method to solve for $T_s$

$T_s$ =350.17 K...=77.17°C

In this particular situation, cat would not be comfortable with walking on a roof since its surface temperature is 77.17°C which may even cause burns to its paws.

4 0

2 years ago

You do a certain amount of work on an object initially at rest, and all the work goes into increasing the object’s speed. If you

EleoNora [17]

Answer:

$\sqrt{2}v$