Answer:
7200 N/m
Explanation:
Metric unit conversion
100g = 0.1 kg
5 cm = 0.05 m
50 cm = 0.5 m
As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground
The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m
Let g = 10 m/s2. The change in potential energy can be calculated as the following:
Therefore, as elastic energy is converted to potential energy and work of friction:
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Explanation:
Bond Enthalpy : It is defined as amount of energy required to break a the particular bond in there gaseous state. It is also known as bond energy. It units are kJ/mol.
- Breaking of a bond is an Endothermic process (energy absorbed from the surroundings).
- Formation of bond is an Exothermic process (energy is released to the surroundings).
If the average bond enthalpy for a C-H bond is 413 kJ/mol, When the C-H bond breaks in which energy will be required ,which will be an endothermic reaction.
Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:
Our values are given by,
As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:
Skate 2:
Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:
Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Answer:
(i) The angular speed of the small metal object is 25.133 rad/s
(ii) The linear speed of the small metal object is 7.54 m/s.
Explanation:
Given;
radius of the circular path, r = 30 cm = 0.3 m
number of revolutions, n = 20
time of motion, t = 5 s
(i) The angular speed of the small metal object is calculated as;
(ii) The linear speed of the small metal object is calculated as;
