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allsm [11]
1 year ago
6

At left A red ball in a box with arrows pointing away from the ball in all directions. In the middle, a blue ball in a box with

arrows pointing toward the ball from all directions. At right, a grey ball in a box. Which object represents a negatively charged particle? Which object represents a positively charged molecule? Which object represents an uncharged molecule? Which object will not move when in an electric field?
Physics
2 answers:
user100 [1]1 year ago
8 0

Answer: The answers are, 1) B. 2) A.  3) C.   4) C.

here is the proof maybe u will understand it more than the numbers and letters lm.ao

Explanation:

MAVERICK [17]1 year ago
6 0

Answer:

first one is b 2nd one is a 3rd is c and the 4th one is c also

Explanation: have a nice day

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3 years ago
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Answer:

1300 m

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 20 s to 70 s is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.

The total distance covered by the train during the entire journey is the area of the speed-time graph.

Area=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)

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As velocity is in m/s and time is in s so the unit of area is m

Hence, the total distance is  1300m.

8 0
3 years ago
Two point charges, 1.8 pC and −1.8 pC, are separated by 7 µm. What is the dipole moment of this pair of charges?
Salsk061 [2.6K]

Answer: 12,600,000Cm

Explanation:

From the data's;

Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C

Distance = 7 micrometer, is equal to 0.0000070m

From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.

Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm

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3 0
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i really wish i can help

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but we are not all einsteins

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