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Mandarinka [93]
3 years ago
11

If a train travels at a constant 18.0 m/s, how far would it move in one hour? In 1.00 minute? In 1.00 second?

Physics
1 answer:
In-s [12.5K]3 years ago
5 0

Explanation:

Distance = speed × time

d = (18.0 m/s) (1 hr × 3600 s/hr)

d = 64,800 m

d = (18.0 m/s) (1 min × 60 s/min)

d = 1080 m

d = (18.0 m/s) (1 s)

d = 18.0 m

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If you drop an object, it accelerates downward at 9.8 m/s2 (in the absence of air resistance). If, instead, you throw it downwar
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*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

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