Specific heat capacity= heat energy/mass×temperature rise
962°C - 20°C = 942K
Heat energy (Eh) = 239 × 1.55 × 942
Eh= 348963.9J
shc of Ag: 238.6 J/kg-K
m of Ag: 1.55kg
Answer:
We define voltage as the amount of potential energy between two points on a circuit. One point has more charge than another. This difference in charge between the two points is called voltage.
Answer:
0.74 s
Explanation:
We can solve the problem by using the following SUVAT equation:
where
d = 5.0 m is the displacement
u = 7.2 m/s is the initial velocity
a = -1.1 m/s^2 is the acceleration (which is negative since it is a deceleration)
t is the time
Substituting numbers into the equation, we find:
This is a second-order equation, whose solutions are given by:
And the solutions are
t = 0.74 s
t = 12.36 s
The solution we are looking for is the first one, because it corresponds to the first time at which the hockey puck has travelled the distance of 5.0 m, reaching the goal.
Mass of Derrick at a height 4.3 m having 2730 joule of energy is 65 Kg.
<h3>What is the expression of gravitational potential energy near the earth surface?</h3>
- Mathematically, gravitational potential energy near the earth surface= m×g×h
- m= mass of Derrick, g= acceleration due to gravity, h= height at which Derrick present
- Then, mass (m) of Derrick = potential energy/ (g×h)
<h3>What is the mass of Derrick, if he gains 2730 joule of energy at 4.3m above the ground?</h3>
Mass of Derrick= 2730/(9.8×4.3)
= 65 kg
Thus, we can conclude that the mass of Derrick is 65 Kg.
Learn more about the gravitational potential energy here:
brainly.com/question/26588957
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Answer :
<em>(b) 4d orbitals would be larger in size than 3d orbitals</em>
<em>(e) 4d orbitals would have more nodes than 3d orbitals</em>
Explanation :
As we move away from one orbital to another, the distance between nucleus and orbital increases. So, 4d orbitals would be far to the nucleus than 3d orbitals.
Hence, 4d orbitals would be larger in size than 3d orbitals.
Number of nodes is any orbital is n - 1 where, n is principal quantum number.
So, number of orbital in 4d is 3.
And number of orbital in 3d is 2.
So, options (b) and (e) are correct.
