Answer:
Explanation:
a )
Pressure on the bottom
= hdg
= 2.5 x 820 x 9.8
= 20090 N / m²
b )
force = area x pressure
= 20090 x 10 x 6 N
= 1205400 N
c )
Force on 10m x 2.5 m face
center of gravity of the tank will lie at height 2 .5 / 2 = 1.25 m
Pressure = hdg
= 1.25 x 820 x 9.8
= 10045 N/m²
force = pressure x area
= 10045 x 10 x 2.5
= 251125 N .
Answer:
F = [M] × [L1 T-2] = M1 L1 T-2.
Explanation:
Therefore, Force is dimensionally represented as M1 L1 T-2.
First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m
Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = <span> 1/2*(625) /(7) = 44.6428 m
The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m
Answer: 14.16
Explanation:
Given
d = 38cm
r = d/2 = 38/2 = 19cm = 0.19m
K.E = 510J
m = 10kg
I = 1/2mr²
I = 1/2*10*0.19²
I = 0.18kgm²
When it has 510J of Kinetic Energy then,
510J = 1/2Iω²
ω² = 1020/I
ω² = 1020/0.18
ω² = 5666.67
ω = √5666.67 = 75.28 rad/s
Velocity is the block, v = ωr
V = 75.28 * 0.19
V = 14.30m/s
The "effective mass" M of the system is
M = (14.0 + ½*10.0) kg = 19.0 kg
The motive force would be
F = ma
F = 14 * 9.8
F = 137.2N
so that the acceleration would be
a = F/m
a = 137.2/19
a = 7.22m/s²
Finally, using equation of motion.
V² = u² + 2as
14.3² = 0 + 2*7.22*s
204.49 = 14.44s
s = 204.49/14.44
s = 14.16m
Answer:
6840 N
Explanation:
The force acting on the car can be found by using Newton's second law:
F = ma
where
F is the net force on the car
m is the mass of the car
a is its acceleration
For the car in this problem,
m = 1800 kg
Substituting,
