Answer:
power =( 890 N x 12 m ) / 22 s=
= 485 Watts
Explanation:
Answer:
W =1562.53 N
Explanation:
It is given that,
Radius of the aluminium ball, r = 24 cm = 0.24 m
The density of Aluminium, 
We need to find the thrust and the force. The mass of the liquid displaced is given by :
V is volume
Weight of the displaced liquid
W = mg
So,
So, the thrust and the force is 1562.53 N.
I believe the correct
form of the energy function is:
u (x) = (3.00 N)
x + (1.00 N / m^2) x^3
or in simpler
terms without the units:
u (x) = 3 x +
x^3
Since the
highest degree is power of 3, therefore there are two roots or solutions of the
equation.
Since we are to
find for the positions x in which the force equal to zero, u (x) = 0,
therefore:
3 x + x^3 = u
(x)
3 x + x^3 = 0
Taking out x:
x (3 + x^2) = 0
So one of the
factors is x = 0.
Finding for the
other two factors, we divide the two sides by x and giving us:
x^2 + 3 = 0
x^2 = - 3
x = sqrt (- 3)
x = - 1.732 i, 1.732
i
The other two
roots are imaginary therefore the force is only equal to zero when the position
is also zero.
Answer:
x = 0
Answer:
10.58 ft
Explanation:
Force, F = 1.4 N
Diameter of membrane = 7.4 mm
radius of membrane, r = 7.4 / 2 = 3.7 mm = 3.7 x 10^-3 m
Area, A = 3.14 x (3.7 x 10^-3)^2 = 4.3 x 10^-5 m^2
Density, d = 1.03 x 10^3 kg/m^3
Pressure at depth, P = h x d x g
Let h be the depth.
Pressure = force / Area
h x 1.03 x 10^3 x 9.8 = 1.4 / (4.3 x 10^-5)
h = 3.225 m = 10.58 ft
Thus, the depth of water is 10.58 ft.
The net force on the box parallel to the plane is
∑ F[para] = mg sin(24°) = ma
where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.
Solve for a :
g sin(24°) = a ≈ 3.99 m/s²
The box starts at rest, so after 7.0 s it attains a speed of
(3.99 m/s²) (7.0 s) ≈ 28 m/s
