Answer:
-13.18°C
Explanation:
To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.
Its definition is given by the function
Where,
Q = The amount of heat transferred
t = time
k = Thermal conductivity constant
A = Cross-sectional area
The difference in temperature between one side of the material and the other
d= thickness of the material
The problem says that there is a loss of heat twice that of the initial state, that is
Replacing,
Solvinf for 
,
Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C
Answer:
8 cm
Ladder is safer than to walk on it
Explanation:
Streched = Final length - initial length
= 66 - 58
= 8 cm
When your are walking along frozen ice pond your weight is not distributed .It's acting on directly on the ice pond because of contact Area is too small between pond and you. There for According to pressure equation ,
P = F/A
When area is less pressure is less.
If use ladder instead of going by foot. Contact area will be high between pond and ladder. There for pressure is reduced .
Answer:
I = 16 kg*m²
Explanation:
Newton's second law for rotation
τ = I * α Formula (1)
where:
τ : It is the moment applied to the body. (Nxm)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Kinematics of the wheel
Equation of circular motion uniformly accelerated :
ωf = ω₀+ α*t Formula (2)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (rad)
Data
ω₀ = 0
ωf = 1.2 rad/s
t = 2 s
Angular acceleration of the wheel
We replace data in the formula (2):
ωf = ω₀+ α*t
1.2= 0+ α*(2)
α*(2) = 1.2
α = 1.2 / 2
α = 0.6 rad/s²
Magnitude of the net torque (τ )
τ = F *R
Where:
F = tangential force (N)
R = radio (m)
τ = 80 N *0.12 m
τ = 9.6 N *m
Rotational inertia of the wheel
We replace data in the formula (1):
τ = I * α
9.6 = I *(0.6
)
I = 9.6 / (0.6
)
I = 16 kg*m²
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Hope this helps! :)
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
