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Shkiper50 [21]
3 years ago
7

You poll students on whether they prefer hotdogs, chicken fingers, salad, or

Physics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

A. Bar Graph

Explanation:

It is easier to see the data.

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3 years ago
A particle has velocity v⃗1 v → 1 as it accelerates from 1 to 2. What is its velocity vector v⃗2 v → 2 as it moves away from poi
Jobisdone [24]

Answer:

The velocity of the particle will be downward.

Explanation:

Given that,

The velocity of a particle is v₁. It is accelerated from 1 to 2.

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We need to find the velocity of particle

According to figure,

A particle moves downward from 1 to 2 with the velocity v₁ and after that the particle moves downward from 2 to 3 with the velocity v₂.

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3 years ago
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A center-seeking force related to acceleration is _______ force
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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}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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2 years ago
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