Answer:
2.068 x 10^6 m / s
Explanation:
radius, r = 5.92 x 10^-11 m
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.
centripetal force = 
Electrostatic force = 
where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2
So, balancing both the forces we get



v = 2.068 x 10^6 m / s
Thus, the speed of the electron is give by 2.068 x 10^6 m / s.
Isothermal Work = PVln(v₂/v₁)
PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J
Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,
1.7 kJ = (5487.24)In(v₂/19)
1700 = (5487.24)In(v₂/19)
In(v₂/19) = (1700/5487.24) = 0.3098
In(v₂/19) = 0.3098
(v₂/19) =

v₂ = 19*

v₂ = 25.8999
v₂ ≈ 26 L Option b.
Statements that are true as regards exposure control plan and its updating are;
<em>Updates must have the reflection of changes in tasks as well in procedures.</em>
<em>Updates must reflect changes in positions that affect occupational exposure.</em>
<em>Updates must have the cost of PPE that is needed and necessary to reduce exposure</em>
An exposure control plan can be regarded as the framework for compliance between the employer and the workers.
- This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.
- This plan gives hope to workers in term of protection when working with their Employer.
- There are some elements that is associated with Exposure Control Plan, and theses are;
- Health hazards as well as risk that is attributed to each product in the worksite.
- Statement of purpose.
- procedures and practices in a written form
- Responsibilities from the Manager, CEO, designated resources and employer.
Therefore, exposure control plan is avenue to protect workers from bloodborne pathogens.
brainly.com/question/1203927?referrer=searchResults
Answer:
Part a)
A = 0.066 m
Part b)
maximum speed = 0.58 m/s
Explanation:
As we know that angular frequency of spring block system is given as

here we know
m = 3.5 kg
k = 270 N/m
now we have


Part a)
Speed of SHM at distance x = 0.020 m from its equilibrium position is given as



Part b)
Maximum speed of SHM at its mean position is given as


Answer:
The gravitational force is 130.
Explanation:
During this problem you have to multiply the 65 and the 0.6.