To solve this problem, we know that:
1 Albert = 88 meters
1 A = 88 m
The first thing we have to do is to square both sides of
the equation:
(1 A)^2 = (88 m)^2
1 A^2 = 7,744 m^2
<span>Since it is given that 1 acre = 4,050 m^2, so to reach
that value, 1st let us divide both sides by 7,744:</span>
1 A^2 / 7,744 = 7,744 m^2 / 7,744
(1 / 7,744) A^2 = 1 m^2
Then we multiply both sides by 4,050.
(4050 / 7744) A^2 = 4050 m^2
0.523 A^2 = 4050 m^2
<span>Therefore 1 acre is equivalent to about 0.52 square
alberts.</span>
Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Answer: 20 kgm/s
Explanation:
Given that M1 = M2 = 10kg
V1 = 5 m/s , V2 = 3 m/s
Since momentum is a vector quantity, the direction of the two object will be taken into consideration.
The magnitude of their combined
momentum before the crash will be:
M1V1 - M2V2
Substitute all the parameters into the formula
10 × 5 - 10 × 3
50 - 30
20 kgm/s
Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s
Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.
