Harry is a professional astronomer. The pictures below show two types of telescopes Harry owns.
1 answer:
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For this case we have that by definition, the momentum is given by:
Where,
- <em>m: mass </em>
- <em>v: speed </em>
Therefore, replacing values we have:
From here, we clear the value of the speed:
Answer:
The magnitude of velocity is:
Answer:
B, C and E
Explanation:
The unit of resistance in the international system is the Ohm, the equation that describes the resistance is:
Where (l) is for lenght of the wire, (S) is the area and (p) its the constant associated to the conductor.
It's related by the Ohm's Law:
Answer:
i) The period of the particle is 0.3 seconds
ii) The angular velocity is approximately 20.94 rad/s
iii) The linear velocity is approximately 1.047 m/s
iv) The centripetal acceleration is approximately 6.98 m/s²
Explanation:
The given parameters are;
The number of revolution of the particle, n = 800 revolution
The time it takes the particle to make 800 revolutions = 4 minutes
The dimension of the circle = 5 cm = 0.05 m
Given that the dimension of the circle is the radius of the circle, we have;
i) The period of the particle, T = The time to complete one revolution
T = 1/(The number of revolutions per second)
∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s
The period, T = 3/10 seconds = 0.3 seconds
ii) The angular velocity, ω = Angle covered/(Time)
800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes
∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3
The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s
iii) The linear velocity, v = r × ω
∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s
iv) The centripetal acceleration, = v²/r
∴ The centripetal acceleration, = (π/3)²/(0.05) = 20·π/9
The centripetal acceleration, = 20·π/9 m/s² ≈ 6.98 m/s²