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babunello [35]
3 years ago
11

The charge imbalance that results from this movement of charge will generate an additional electric field in the region within t

he rod In what direction will this field point? A. It will point to the right and enhance the initial applied field B. It will point to the left and oppose the initial applied field
Physics
1 answer:
deff fn [24]3 years ago
8 0

Answer:

B. It will point to the left and oppose the initial applied field

Explanation:

Let the charge be positive and  and electric field be towards right . Charge will move towards right in the field due to which positive charge will accumulate towards right and excess of positive charge will lie there . This new charge distribution will create a field towards the left which is opposite to external electric field.

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3 years ago
Please need help with this
NeTakaya

<u>Note that</u>:

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3 years ago
A car travels at a constant rate for 25 miles, going due east for one hour. Then it travels at a constant rate another 60 miles
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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
Len [333]

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

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}

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

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3 years ago
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