Answer:
Work done = 13605.44
Explanation:
Data provided in the question:
For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J
The relation between Energy (U) and the elongation (s) is given as:
U = 
................(1)
where,
k is the spring constant
on substituting the valeus in the above equation, we get
3.0 = 
or
k = 13605.44 N/m
now
for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m
using the equation 1, we have
U = 
or
U = 26.149 J
Also,
Work done = change in energy
or
W = 26.149 - 3.0 = 23.149 J
The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
Therefore, option A is correct option.
Given,
Mass m = 14g
Volume= 3.5L
Temperature T= 75+273 = 348 K
Molar mass of CO = 28g/mol
Universal gas constant R= 0.082057L
Number of moles in 14 g of CO is
n= mass/ molar mass
= 14/28
= 0.5 mol
As we know that
PV= nRT
P × 3.5 = 0.5 × 0.082057 × 348
P × 3.5 = 14.277
P = 14.277/3.5
P = 4.0794 atm
P = 4.1 atm.
Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
learn more about pressure:
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Answer:
The amount of work the factory worker must to stop the rolling ramp is 294 joules
Explanation:
The object rolling down the frictionless ramp has the following parameters;
The mass of the object = 10 kg
The height from which the object is rolled = 3 meters
The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp
Where;
The potential energy P.E. = m × g × h
m = The mass of the ramp = 10 kg
g = The acceleration due to gravity = 9.8 m/s²
h = The height from which the object rolls down = 3 m
Therefore, we have;
P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules
The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules
U have to *modify it to increase its ground clearance*
Answer:
Acceleration, 
Explanation:
Initial speed of the skater, u = 8.4 m/s
Final speed of the skater, v = 6.5 m/s
It hits a 5.7 m wide patch of rough ice, s = 5.7 m
We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :
So, the acceleration on the rough ice 
and negative sign shows deceleration.
