Answer:
13.524 N
Explanation:
Volume and densities are given as:
ρ1 = 2.6 g/cm³ => 2600 kg/m³ ; V1 = 0.50 L => 0.5 x 10^-3 m³
ρ2 = 1.0 g/cm³ => 1000 kg/m³ ; V2= 0.25 L => 0.25 x 10^-3 m³
ρ3 = 0.7 g/cm³ => 700 kg/m³ ; V3 = 0.4 L => 0.4 x 10^-3 m³
Next is to calculate force exerted on the bottom of the container due to these liquids:
F= ρ1V1g + ρ2 V2 g+ ρ 3 V3g
where ,
ρ= density
V= volume
g= 9.8m/s²
F= g( 2600 x 0.5 x 10^-3 + 1000 x 0.25 x 10^-3 + 700 x 0.4 x 10^-3)
F= 9.8 (1.38)
F= 13.524 N
Therefore, the force on the bottom of the container due to these liquids is 13.524 N
Explanation:
S=(V^2-U^2)/2a a=g (gravity) a=10
=(0^2-25^2/2*(-10)
=625/20
=31.25m
Answer:
(a) Distance traveled = 75.3846 m
(b) Velocity of car at that instant will be 14 m/sec
Explanation:
We have given acceleration of the car 
Initial velocity of the cart u = 0 m/sec
(a) According to second equation of motion we know that 
So distance traveled by car 
As the truck is moving with constant speed
So distance traveled by truck 
As the truck overtakes the car
So 
So distance traveled 
(b) From second equation of motion we know that v = u+at
So v = 0+1.3×10.769 = 14 m /sec
B) O False because that’s just what I wanted to put that’s what I think the answer is
