Answer:
F=9/5(°C)+32
Explanation:
Lets take an example
- Take 25°C
A meter stick is balanced at the 50 cm mark. You tie a 20 N weight at the 20 cm mark. Where should a 30 N weight be placed so th
1 answer:
Answer:
20 cm to the right of the center or 20+50 = 70 cm from the left side.
Explanation:
The length of meter stick is 1 m = 100 cm
Balance point on 50 cm
From the center the 20 N weight is 50-20 = 30 cm
Torque is obtained when force is multiplied with the distance
As the force is conserved we have
The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.
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Answer: a) speed = 3.45 × 10^-2 m/s
b) speed = 1.38 × 10^-1 m/s
Explanation: shown in the attachment
Answer:
F = 3.86 x 10⁻⁶ N
Explanation:
First, we will find the distance between the two particles:
where,
r = distance between the particles = ?
(x₁, y₁, z₁) = (2, 5, 1)
(x₂, y₂, z₂) = (3, 2, 3)
Therefore,
Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:
where,
F = magnitude of force = ?
k = Coulomb's Constant = 9 x 10⁹ Nm²/C²
q₁ = magnitude of first charge = 2 x 10⁻⁸ C
q₂ = magnitude of second charge = 3 x 10⁻⁷ C
r = distance between the charges = 3.741 m
Therefore,
<u>F = 3.86 x 10⁻⁶ N</u>