The mass of the water is 14.9 g
Explanation:
When a certain amount of a susbtance is heated, the temperature of the substance increases according to the equation
where
Q is the amount of energy supplied to the substance
m is the mass of the substance
is its specific heat capacity
is the change in temperature
In this problem, we have:
Q = 2500 J of energy supplied to the water
is the specific heat capacity of water
is the change in temperature of the water
Therefore we can solve for m to find the mass of the water:
Learn more about specific heat capacity:
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Answer:
<h3>The 28 loops wound on the square armature</h3>
Explanation:
Peak output voltage 
V
Area of square armature 
Magnetic field 
T
Angular frequency 
According to the law of electromagnetic induction,
Where 
number of loops of wire.
≅ 28
Thus, 28 loops of wire should be wound on the square armature.
Answer:
electric field E = (1 /3 e₀) ρ r
Explanation:
For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.
The charge within our surface is
ρ = Q / V
Q ’= ρ V
'
The volume of the sphere is V = 4/3 π r³
Q ’= ρ 4/3 π r³
The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area
I E da = Q ’/ ε₀
E A = E 4 πi r² = Q ’/ ε₀
E = (1/4 π ε₀) Q ’/ r²
Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant
R = Q ’/ V’ = Q / V
How you want the solution depending on the density (ρ) and the inner radius (r)
Q ’= R V’
Q ’= ρ 4/3 π r³
E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³
E = (1 /3 e₀) ρ r
Answer:
Option C
Explanation:
Given that
Motor force is 250 N
Force of friction is 750 N
Weight is 8500 N
And, the normal force is 8500 N
Now based on the above information
Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force
