Result: -50.005 kJ
Details:
Provided Data
mass of the system = 10 kg
work done = 0.147 kJ/kg
Elevation change 
initial speed 
Final Speed 
Specific internal Energy 
according to the first Law of thermodynamics
where KE represents kinetic energy
PE indicates potential energy
U denotes internal Energy
Q = 1.47 + 3.375 - 4.850 - 50
Q = -50.005 kJ
Refer to the diagram below.
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
In the study of physics, Hooke's law can be expressed as:
F = kx
This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.
In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.
Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.
Answer:
h = 12.8 cm
Explanation:
The initial parameters are as follows:
distance = 6.4 cm
- when the object descends, its weight matches the spring's force
weight = spring force
mg = ky... equation 1
- potential energy stored in a stretched spring = work done by the spring
mgh = 0.5 x k x h^{2}....equation 2
- Substituting from equation 1 into equation 2
kyh = 0.5 x k x h^{2}
y = 0.5 x h
2y = h
- where y is 6.4, yielding the maximum elongation as
h = 2 x 6.4 = 12.8 cm
Answer:
Explanation:
Amount of gold deposited = 0.5 g
Gold's molar mass = 197 g/mol
Time duration, t = 6 hours
= 6 × 3600
= 12600 s
Calculation of moles: mass/molar mass
= 0.5/197
= 0.00254 mole
Assuming
Au --> Au+ + e-
Faraday's constant = 9.65 x 10^4 C mol-1
Charge, Q = 96500 × 0.00254
= 244.924 C
Relation: Q = I × t
Thus, I = 244.924/12600
= 0.011 A
= 11.34 mA.
