Responder:
Fb = 17,5 kN
Explicación:
La expresión de la fuerza de empuje
Fb = ρgV ---------- 1
dónde
Fb = fuerza de flotación
ρ = densidad del fluido
g = aceleración debida a la gravedad
V = volumen de fluido
Paso uno:
datos dados
radio del cilindro = 0,3 m
altura h = 8m
densidad = 790 kg / m ^ 3
g = 9,8 m / s ^ 2
El volumen del cilindro se expresa como
V = πr ^ 2h
V = 3,142 * 0,3 ^ 2 * 8
V = 3,142 * 0,09 * 8
V = 2,26
El volumen es 2.26m ^ 3, el volumen del cilindro es igual al volumen de alcohol disparado
La expresión de la fuerza de empuje es
Fb = ρgV
sustituyendo tenemos
Fb = 790 * 9,8 * 2,26
Fb = 17514,26N
Fb = 17,5 kN
Answer:
2.5 cm
40 D
Explanation:
When the radius of curvature of a lens is divided by 2 we get the focal length of the lens.
Focal length is given by
The focal length of the lens is 2.5 cm
When we divide 1 by the focal length in the unit of meters we get the power of a lens
Power of a lens is given by
The power of the lens is 40 D
Answer:
Explanation:
Speed is a measurment that tells him much distance a body travels over a certain amount of time (usually measured in meters per second (m/c))
<h2>Correct answer:</h2>
<h2>Explanation:</h2>
We can use voltage divider to solve this problem that is defined as the passive linear circuit producing an output voltage 
that is a fraction of its input voltage 
. So we can use the formula:
Answer:
A. The electron will begin to move along the axis, towards the centre and the instantaneous velocity because the force acting on it depends largely on acceleration and x until it reaches maximum velocity at centre.
B. Veloctiy (Vb) = 1.66m/s
Explanation:
Given the following data
x(a) = 0.3m
x(b) = 0
q = 1.6×10^-19
Q = 24nc
r = 0.15m
Required: the motion of the electron and the velocity (Vb)
1. At point A the electron will begin to move along the axis from point A to point B, the magnitude of the electric field will change while moving which depends on that and this will produce instantaneous force which will later change and the acceleration will change too while moving, the velocity would reach maximum value at point B
2. Potential energy and kinetic energy are given by
U(a) + K(a) = U(b) + K(b). . .1
Initial P.E and K.E are given as
U(a) = kQ/√x²(a) + a2
By substitution, we have
U(a) = 9×10^9 × (-1.9×10^-19)×24×10^-9/√(0.15)²+(0.3²)
U(a) = -1.03×10^-16
Final P.E and K.E are given as
U(b) = KQ/√x²(b) + a2
By substitution, we have
U(b) = 9×10^9×(-1.9×10^-19)×24×10^-9/√(0.15)²+(0)²
U(b) = -2.3×10^16
3. By substitution into equation 1 becomes
-1.03×10^-6 - 2.3×10^-16 + MV²(b)/2
V(b) = √2×1.27×10^-16/9.1×10^31
V(b) = 1.66×10^7m/s
