<span>equal-arm balance = 12 kg
spring scale = 2 kg
The equal arm balance measures the mass of an object by using a counter mass on the opposite plate of the balance. The force of gravity affects both the mass being tested and the mass standards being compared against equally. So if the local gravitational field changes, those changes affect both the tested mass and the standard masses equally.
Contrast that to the spring scale. In that scale, the spring provides a calibrated level of force irrespective of the local gravitational field. So if the local gravity is higher, the force indicated also is higher. And if the gravitational field is lower, the indicated force is also lower. The strength of the spring DOES NOT CHANGE with changes in the local gravitational field.</span>
<h2>Answer</h2>
Radius is 421 m.
<u>Explanation </u>
A car with a mass of 1,500 kg requires a centripetal force of 640 N to safely follow a circular curve in the road at 13.4 m/s. Therefore, for radius we use formula which is Fc = mv ^ 2 / r,
As mass = m = 1500kg,
Centripetal force = Fc = 640N,
Velocity = v = 13.4 m / s
By putting values, Fc = mv ^ 2 / r,
r = mv ^ 2 / Fc,
=> r = ( 1500kg ) . ( 13.4 m / s) ^ 2 / 640,
=> r = ( 1500kg ) . ( 179.56 ) / 640,
r = 269340 / 640,
=> r = 420.84 m.
Radius is 421 m.
If you drop a ball off a building it increases its speed by 10 m/s so in 3 seconds it’s speed would be 30 m/s and it’s distance would be 45m
