200. newton·meters of work is put into a machine over a distance of 20. meters. The machine does 150. newton·meters of work as i
1 answer:
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Missing figure and missing details can be found here:
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Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
points in the opposite direction of the displacement of the cart, and where
therefore, the work done by the weight is
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
therefore, the work done by the weight is
Answer:
Explanation:
Electric field due to a point charge Q at a point at distance d is given by the relation
E =
Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.
The charges are situated on the corners of a square in such a way that
equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.
On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero
Overall, net field due to all the four charges will be zero