Answer:
A force is a push or pull upon an object resulting from the objects interaction with another object
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.
If the machine is 100% efficient, then its
Mechanical Advantage is (450/50) = 9 .
If the machine is less than 100% efficient,
then the MA is more than 9 .
Answer:
6.53 m/s²
Explanation:
Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.
Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:
T = m₁a (1)
m₂g - T = m₂a (2)
substituting T = m₁a in equation 2:
m₂g - m₁a = m₂a
m₂a + m₁a = m₂g
a(m₁ + m₂) = m₂g
a = m₂g / (m₁ + m₂)
a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²
Both objects have an acceleration of 6.53 m/s²
Answer:
The efficiency of this engine is 37.63 %.
Explanation:
Given;
useful output work done by the combustion engine, = 356 kJ
input work, = 946 kJ
The efficiency of the combustion engine is calculated as;
Therefore, the efficiency of this engine is 37.63 %.
