How is distance related to force in this experiment to mass?
1 answer:
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Answer:
The final temperature of the element = 262.67°C
The power dissipated in the heating element initially = 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W
Explanation:
Our given parameters include;
A Nichrome heating element operates on 120 V.
Voltage (V) = 120V
Initial Current (I₁) = 1.36 A
Initial Temperature (T₁) = 28°C
Final Current (I₂) = 1.23 A
Final Temperature (T₂) = unknown ????
Temperature dependencies of resistance is given by:
---------------------- (1)
in which R₁ is the resistance at temperature T₁
is the resistance at temperature T₂
Given that V= IR
R =
Therefore, the resistance at temperature 28°C is;
= 88.24Ω
=
= 97.56Ω
From (1) above;
97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]
1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)
0.1056 = 4.5×10⁻⁴(T₂-28°C)
T - 28° C = 234.67
T = 234.67 + 28° C
T = 262.67 ° C
(b)
What is the power dissipated in the heating element initially and when the current reaches 1.23 A
The power dissipated in the heating element initially can be calculated as:
P = I²₁R₂₈
P = (1.36A)²(88.24Ω)
P = 163.209 W
P ≅ 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:
P = (1.23)²(97.56Ω)
P = 147.598524
P ≅ 147.60 W
Answer:
1) The mass of the student is approximately 73.39 kg
2) The net force on the student is approximately 947.523 N
3) The value the scale will read is approximately 96.59 kg
Explanation:
The given parameters are;
The weight of the student = 720 N
The speed at which the elevator is decreasing = 3.1 m/s²
1) The weight of the student = The mass of the student × The acceleration due to gravity
The acceleration due to gravity is a constant = 9.81 m/s²
Substituting the known values gives;
720 N = The mass of the student × 9.81 m/s²
∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg
2) The forces acting on the student are;
i) The force of gravity which is the weight of the student acting downwards
ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student
The net force, = The weight of the student + The inertia force of the slowing elevator
∴ The net force, = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N
3) The scale will read the mass of the student as follows;
Mass reading of student on the scale = Force on scale/9.81
∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg
The value the scale will read = 96.59 kg.