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Cottage and Small Industries Development Committee to organize National Industrial Goods and Technology Expo
By Glocal Khabar - 2368 0
National Industrial Goods and Technology Expo- Glocal Khabar
Kathmandu, February 19, 2018: Cottage and Small Industries Development Committee is set to organize the twenty-ninth edition of National Industrial Goods and Technology Expo from March 23, 2018 at Bhrikutimandap, Kathmandu.
The motto of the five-day event is ‘Let’s use home-made products, move ahead towards prosperity.’ The committee has shared that handicrafts, wool and bamboo products, goods made from handmade papers, different types of pickles and Palpali Dhaka would be put on display in the expo.
Figure A shows cross section of a land form or rock. In Figure B, compression stress is applied on it. When compression stresses are applied on a rock, it squeezes the rock cause fold or fracture. The fault formed by compression stress is called thrust fault. If the compression stresses/ force continue to act on a rock it will converge and form thrust fault. In Figure C, tension stresses is applied on the rock. When a tension stress applied on a rock it deforms/ lengthen. There are three type of deformations occur due to tension stresses. One is elastic deformation, in which, rock retains it original shape when force/stresses are removed. Second is plastic deformation, in which rock lengthen and change occur permanently. Third type of deformation is result into fracture or breaking of rock. In Figure C, shear stresses are applied on rock. Shear stresses are applied with equal magnitude but in opposite direction. It cause breaking of rock.
That's called the "normal" to the surface at that point.
Answer:
Option E is correct 310N
Explanation:
Given that the force used to push the crate is F = 200N
The force directed 20° below the horizontal
Mass of crate is m = 25kg
Weight of the crate can be determine using
W = mg
g is gravitational constant =9.8m/s²
W = 25×9.8
W = 245 N
Check attachment. For free body diagram and better understanding
Using newton second law along the vertical axis since we want to find the normal force
ΣFy = m•ay
ay = 0, since the body is not moving in the vertical or y direction
N—W—F•Sin20 = 0
N = W+F•Sin20
N = 245+ 200Sin20
N = 245 + 68.4
N = 313.4 N
The normal force is approximately 310 N to the nearest ten
