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Sati [7]
3 years ago
5

A test charge of -1.4 × 10-7 coulombs experiences a force of 5.4 × 10-1 newtons. Calculate the magnitude of the electric field c

reated by the negative test charge.
A. 1.4 × 106 newtons/coulomb
B. 1.9 × 106 newtons/coulomb
C. 5.4 × 10-1 newtons/coulomb
D. 3.6 × 106 newtons/coulomb
Physics
2 answers:
kotykmax [81]3 years ago
8 0

Answer: D. 3.6*10^-6

Explanation:

Electric field strength E= Force on a charge / charge

E = F/q

E = 5.4*10^-1 / 1.4*10^-7

E = 3.857*10^6 N/C

mr_godi [17]3 years ago
3 0
I would say C 5.4 x 10-1 Newtons/coulomb
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                                          d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}

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                                             d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}.

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                                              t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}.

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

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Therefore, the average speed of the car is 50 km/hr.

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