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seraphim [82]
3 years ago
9

In the southern hemisphere, the longest day of the year happens during _____.

Physics
2 answers:
Luden [163]3 years ago
8 0
I think it is June. June is the longest day/
blsea [12.9K]3 years ago
7 0
It is June. June 20th
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When you approach a yield sign, while trying to enter or merge onto another road, traffic already on that road has the right of
kogti [31]
TRUE. When you approach a yield sign, while trying to enter or merge onto another road, traffic already on that road has the right of way.
5 0
3 years ago
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A jet liner must reach a speed of 82 m/s for takeoff. If the
SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial (v_{o}) and final speeds (v_{f}), measured in meters per second, of the aircraft and likewise the runway length (d), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed (a), measured in meters per square second:

a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 82\,\frac{m}{s} and d = 1500\,m, then the acceleration that the jet must have is:

a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}

a = 2.241\,\frac{m}{s^{2}}

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0
2 years ago
What is a short circuit?
Marizza181 [45]

Answer:

A short circuit is an electrical circuit that allows a current to travel along an unintended path. It has no or very low electrical impedances. The opposite of a short circuit

Explanation:

3 0
3 years ago
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13 points and brainlyest if possible. Thanks.
nikdorinn [45]
Most likely it would be C not completely sure 
3 0
3 years ago
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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.

4 0
2 years ago
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