The tension on the wire is 52.02 N.
From the question, we have
Density of aluminum = 2700 kg/m3
Area,
A = πd²/4
A = π x (4.6 x 10⁻³)²/4
A = 1.66 x 10⁻⁵ m²
μ = Mass per unit length of the wire
μ = ρA
μ = 2700 kg/m³ x 1.66 x 10⁻⁵ m²
μ = 0.045 kg/m
Tension on the wire = √T/μ
34 = √T/0.045
34² = T/0.045
T = 52.02 N
The tension on the wire is 52.02 N.
Complete question:
The density of aluminum is 2700 kg/m3. If transverse waves propagate at 34 m/s in a 4.6-mm diameter aluminum wire, what is the tension on the wire.
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In a series circuit the total current is the same throughout resistors and so:
The voltage is distributed throughout the resistors and so:
and the total resistance can be calculated by adding up the resistors resistance:
First thing is to calculate the total resistance and so:
And by Omh's law V=IR we have:
And so the total current of the circuit is 1.2 amps i.e. 1.2 A.
B
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Answer:
D) the second at the doorknob
Explanation:
The torque exerted by a force is given by:
where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation
is the angle between the direction of the force and d
In this problem, we have:
- Two forces of equal magnitude F
- Both forces are perpendicular to the door, so
- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force
--> therefore, the 2nd force exerts a greater torque
Answer:
π/10 rads
Explanation:
It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of
2πl rad.
Now, number of periods of 3 minutes in an hour is;
Number of periods = 60/3 = 20 periods
Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.
Thus;
Angular displacement = (1/20) * 2π = π/10 rads