The Professor's centripetal acceleration is 0.044 m/s²
Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.
It is given by:
a = v²/r
where v is the velocity and r is the radius.
Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:
a = v²/r = 0.419²/4 = 0.044 m/s²
The Professor's centripetal acceleration is 0.044 m/s²
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Answer: True.
Explanation:
You would be able to visualize the basketballs height going up and when it sinks down into the hoop.
Answer;
-Sensors
-Sensors are placed on dangerous machinery to detect motion, light, heat, pressure, or another stimulus. Their presence helps protect operators from injury while working on machines.
Explanation;
-Machinery, safety and factory floor sensors and switches help workers become more productive, efficient, and safe.
-Hazardous machines and systems are frequently equipped with safety elements (safety doors) with a locking mechanism to protect the operator. Their function is to prevent hazardous machine functions if the safety door is not closed and locked and to keep the safety door closed and locked until the risk of injury has passed.
Answer:
F = 8.6 10⁻¹² N
Explanation:
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Em₀ = U = q ΔV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v²
Em₀ = Emf
e ΔV = ½ m v²
v =√ 2 e ΔV / m
v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)
v = √(1.8075 10¹⁶)
v = 1,344 10⁸ m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Emo = U = q DV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v2
Emo = Emf
.e DV = ½ m v2
.v = RA 2 e DV / m
.v = RA (2 1.6 10-19 51400 / 9.1 10-31)
.v = RA (1.8075 10 16)
.v = 1,344 108 m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10-19 1,344 108 0.4
F = 8.6 10-12 N
