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Answer: Yes
Explanation: A molecule has a bond dipole but not have a molecular dipole.
Dipole moment: dipole moment is the product of the distribution of the charge of the molecules and the distance between of the molecules.
Example:
lets take an example of molecule in which two electro negative oxygen atoms are attached with Carbon atom. Oxygen being electro negative will attract the shared pair electrons towards itself and partial negative charge will create on oxygen atom and partial positive charge on carbon atom C.
due to formation of partial positive and partial negative charges dipole moment is created between oxygen and Carbon bond on both the sides in opposite direction.
Since, the dipole moment acting in opposite direction the net dipole moment in the molecule is zero.
Hence, Yes, a molecule has bond dipoles but not have a molecular dipole.
Answer:
d. Both A & C
Explanation:
- The velocity of the ball is a vector, whose magnitude indicates its rate of change of position, and it has a direction. In this case, the ball is moving upward, therefore the direction of motion is upward, so the direction of the velocity is upward as well.
- The acceleration of the ball is a vector, whose magnitude indicates the rate of change of the velocity. The direction of the acceleration is:
-- positive if the the magnitude of the velocity is increasing
-- negative if the magnitude of the velocity is decreasing
For a ball thrown upward, the acceleration is given by the acceleration of gravity, . This acceleration points downward, and it is constant during the entire motion. In particular, it does not change direction, as it is always directed downward. Therefore, the acceleration of the ball is downward.
So, the correct answer is
d.Both A & C
Since A and C are both true:
a.The velocity vector is directed upward.
c.The acceleration is directed downward.
Answer:
(i)The current flow in black wire = 9.67 A (ii) The current low in the red wire is 9.68 A (iii) The current flow in neutral wire is 15.36 A (iv) when 240 volt were disconnected current in black wire is 7.68 A (v) when 240 volt were disconnected current in red wire is 7.68 A (vi) 15.36 A (vii) 6.34 (viii) 9.68 A (ix) 12.02 A
Explanation:
Solution
The current drawn by one amp is
I =P/V
I =200/120
I= 1.67 A
(i) The current flow in the black wire is
IBK = 4 * 1.67 A + 1A + 2A
IBK = 9.67 A
(ii) Current flow in the red wire is
IRD = 3 * 1.67 A + 1.67 A + 1A + 2A
= 8.68A + 1 A = 9.68 A
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