Answer:
Kindly check explanation
Explanation:
Given the data:
Trial V (volts) I (mA)
1 1.00 7.2
2 2.10 14.0
3 3.10 20.7
4 4.00 27.2
5 4.90 32.2
Slope = Rise / Run
Rise = y2 - y1 = 32.2 - 7.2 = 25
Run = x2 - x1 = 4.9 - 1.0 = 3.9
Slope = 25 / 3.9 = 6.410
y = mx + c
The intercept, c
Take the point ( 1; 7.2)
Put x = 0
7.2 = 6.410(1) + C
7.2 - 6.410 = C
C = 0.79
An example of a balanced force would be a book sitting on a shelf untouched.
Isaac Newton’s First Law of Motion states that an object at motion stays in motion, and an object at rest stays at rest until acted on by an unbalanced force. A book sitting still is an example of a balanced force because nothing is acting on it; its potential energy is stored while it’s at rest. For this book to become an unbalanced force, an outside force would have to occur (i.e pushing the book or dropping it) that causes it to not be in a state of stillness.
Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m