3 A vector of magnitude 20 is added to a vector of magnitude 25. The magnitude o

Question

3 A vector of magnitude 20 is added to a vector of magnitude 25. The magnitude of this sum can be 4 An airplane makes a gradual 90 turn while flying at a constant speed of 200 m/s. The process takes 20.0 seconds to complete. For this turn the magnitude of the average acceleration of the plane is: 5 A 2.0-kg stone is tied to a 0.50-m string and swung around a circle at a constant angular velocity of 12 rad/s. The torque on the stone about the center of the circle is: 6. The approximate value of g (in m/s^2) at an altitude above Earth equal to one Earth radius is: 7 A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic friction is 0.40. After one end of the plank is raised so the plank makes an angle of 25 with the horizontal, the force of friction is:

Explanation / Answer

3. vector A have magnitude 20 , then vector A = 4.472i

Vector B have magnitude 25, then vector B = 5i

Vector (A + B) = 9.472 i

magnitude | A + B| = 3.02

Hence option B is correct

4. The average acceleration

a = v2 / r

time period T = @ * r / v ( where @ = 90 = pi / 2)

==> r = T * v / @ = 2 *20 * 200 / 3.14 = 2547.77 m

==> a = 200 * 200 / 2547.77

= 15.7 m/s2

I hope that option D will be correct.

5. Torque acting on the stone is

T = mv2 / r

= mr w2

= 2 * 0.5 * 122

= 144 N at the end of the circle.

But at the center of the circle , the radius of the circle is zero. Hence torque acting on the stone is zero.

Correct option is (A)

6. According to inverse square law

acceleration is inversly proportional to square of the distance between objects.

Here the distance from the center of the Earth is 2r

==> a / g = (r / 2r)2 = 0.25

==> a = 0.25 * 9.8 = 2.45 m/s2 ~ 2.5 m/s2

option C is correct

7. Ffr = uk m g cos@ = 0.40 * 5 * 9.8 * cos25 = 17.76 ~ 18 N

option B is correct

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