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amm1812
3 years ago
14

A dipole of moment 0.50 e · nm is placed in a uniform electric field with a magnitude of 1.6 105 N/C. What is the magnitude of t

he torque on the dipole for each of the following situations?
(a) the dipole is aligned with the electric field
(b) the dipole is transverse to (perpendicular to) the electric field
(c) the dipole makes an angle of 26° with the direction of the electric field
(d) Defining the potential energy to be zero when the dipole is transverse to the electric field, find the potential energy of the dipole in the electric field for the orientations specified in Parts (a) and (c).
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0
Too much to read bro bro
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Explanation:

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A train increases its speed steadily from 10 m/s to 20 m/s in
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Answer:

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10+20=30

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3 years ago
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
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Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

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v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

8 0
3 years ago
A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is
pychu [463]

Answer:

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Explanation:

Given;

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Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

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