Answer:
v = 10.84 m/s
Explanation:
using the equation of motion:
v^2 = (v0)^2 + 2×a(r - r0)
<em>due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.</em>
v^2 = 2×g(r - r0)
v = \sqrt{2×(-9.8)×(4 - 10)}
= 10.84 m/s
therefore, the velocity at r = 4 meters is 10.84 m/s
Answer:
The total distance, side to side, that the top of the building moves during such an oscillation = 31 cm
Explanation:
Let the total side to side motion be 2A. Where A is maximum acceleration.
Now, we know know that equation for maximum acceleration is;
A = α(max) / [(2πf)^(2)]
So 2A = 2[α(max) / [(2πf)^(2)] ]
α(max) = (0.025 x 9.81) while frequency(f) from the question is 0.2Hz.
Therefore 2A = 2 [(0.025 x 9.81) / [((2π(0.2)) ^(2)] ] = 2( 0.245 / 1.58) = 0.31m or 31cm
Answer:
Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?
Search instead for Two point charge q, seperated by 1.5cm have change value of +2.0 and -4.0N/C respectively what is the magnitude of the Electric force midway between them?
Answer:
Potential energy =mass* acceleration due to gravity * height
mass*acceleration due to gravity =weight
hence potential energy of the puppy= weight * height
=18*2
=36 joule
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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