The distance between two objects is doubled, the force between them drops to a quarter (1/22<span> or 1/4). If the distance is increased ten times, the force between the two objects decreases to a hundredth of what it was (1/10</span>2<span> or 1/100).</span>
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Answer:
Velocity of Afrom B=21m/s
Acceleration of A from B=1.68m/s°2
Explanation:
Given
Radius r=150m
Velocity of a Va= 54km/hr
Va=54*1000/3600=15m/s
Velocity of b Vb=82km/hr
VB=81*1000/3600=22.5mls
The velocity of Car A as observed from B is VBA
VB= VA+VBA
Resolving the vector into X and Y components
For X component= 15cos60=7.5m/s
Y component=22 5sin60=19.48m/s
VBA= √(X^2+Y^2)
VBA= ✓(7.5^2+19.48^2)=21m/s
For acceleration of A observed from B
A=VA^2/r= 15^2/150=1.5m/s
Resolving into Xcomponent=1.5cos60=0.75m/s
Y component=3cos60=1.5
Acceleration BA=√(0.75^2+1.5^2)
1.68m/s
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Answer:
C)0.59 N
Explanation:
By the Achemides principle we know that the buoyancy force on an object which is immersed in a incompressible fluid at rest is equal to the weight of the fluid displaced. So all we have to do is to find the mass of 60ml.
Density = mass / volume
Mass = density × volume
= 1 g/cm³× 60 cm³
{ as 1ml=1cm³}
= 60 g
So force equals to,
Force = weight = mass × gravitational acceleration
= 0.06×9.8 = 0.59 N