Answer:
There is 148.35 Joules of heat is released in the process.
Explanation:
Given that,
Heat capacity of the object, 
Initial temperature, 
Final temperature, 
We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :
Let the mass of the object is 10 g or 0.01 kg
So,
Q = 148.35 Joules
So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.
Answer:
E = 1.25×10¹³ N/m²
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Given:
F = 10 kg × 9.8 m/s² = 98 N
L = 1 m
dL = 10⁻⁵ m
A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²
Solve:
E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)
E = 1.25×10¹³ N/m²
Round as needed.
D. Teach the public energy conservation
Answer:
E = k*Q₁/R₁² V/m
V = k*Q₁/R₁ Volt
Explanation:
Given:
- Charge distributed on the sphere is Q₁
- The radius of sphere is R₁
- The electric potential at infinity is 0
Find:
What is the electric field at the surface of the sphere?E.
What is the electric potential at the surface of the sphere?V
Solution:
- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.
- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F = k*Q₁/R₁²
- Then the electric field at that point is
E = F/1
E = k*Q₁/R₁² V/m
- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V = k*Q₁/R₁ Volt
