A 4.55 nF parallel-plate capacitor contains 27.5 μJ of stored energy. By how many volts would you have to increase this potentia
1 answer:
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Answer:
1. 3
2. 1.5
3. 3 seconds
4. 0
5. 2.2 seconds
Explanation:
(1)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds
a =
(2)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
Substituting u=10m/s, v=22m/s and t=8 seconds
a =
(3)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making t the subject we have
Substituting u=0m/s since at rest, v=15m/s and a=5
=
(4)
When initial and final velocity are constant, there’s no acceleration as proven below
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds
a =
(5)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making t the subject we have
Substituting u=9m/s since at rest, v=0m/s and a=-4.1
=
During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.
<u>Explanation: </u>
As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.
In first step, the clown runs 7.7 m in north direction, so the image will be as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.
So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.
By using Pythagoras theorem, the total displacement can be found as
Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.
Thus, the total displacement covered by the clown is
Thus, the total displacement by the clown is 27 m.
