Answer
given,
height of the dam = 15 m
effective area of water = 2.3 x 10⁻³ m²
Using energy conservation
v = 17.15 m/s
discharge of water
Q = A V
Q = 2.3 x 10⁻³ x 17.15
Q = 0.039 m³/s
The strength of the gravitational field of a source mass can be measured by the magnitude of the acceleration due to gravity at
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The electrical potential energy of a charge q located at a point at potential V is given by
Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be
In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be
In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
Answer:
1) The radius of the tire is approximately 0.28966 meters
2) The centripetal force is the force that keeps a body moving on a circular path
Explanation:
1) The linear speed of the automobile tire = 20.0 m/s
The period with which the tire turns = 0.091 s
The period = The time it takes to make a complete turn
Therefore;
The number of turns in 1 second = 1/0.091 ≈ 10.989 turns
The distance covered with 10.989 turns, assuming no friction = 10.989 × The circumference of the tire
∴ The distance covered with 10.989 turns, assuming no friction = 10.989 × 2 × π × Radius of the tire
From the speed of the car, 20.0 m/s, we have;
The distance covered in 1 second = 20.0 meters
Therefore;
10.989 × 2 × π × Radius of the tire = 20.0 meters
Radius of the tire = (20.0 meters)/(10.289 × 2 × π) ≈ 0.28966 meters
The radius of the tire ≈ 0.28966 meters
2) The centripetal force is the force required to maintain the curved motion of an object, and having a direction towards the center of the rotary motion.
The centripetal force is given by the formula,
Where;
F = The centripetal force
m = The mass of the object
v = The linear velocity of the object
r = The radius of the rotational motion.
Answer:
So the acceleration of the child will be
Explanation:
We have given angular speed of the child
Radius r = 4.65 m
Angular acceleration
We know that linear velocity is given by
We know that radial acceleration is given by
Tangential acceleration is given by
So total acceleration will be