Which of the following is not a

Iron is essential micro-nutrient for the growth of plant but it is required only in small amount. Iron is required when plant produces chlorophyll and this gives plants oxygen and healthy green color but if a plant is suffering from iron deficiency then the plant's growth becomes stunted and leaves color changes to yellow. For plants requirement of iron is just 1 to 1.5 lb per acre whereas, requirement of nitrogen is 75 to 200 lb per acre.

4 0

3 years ago

A 0.42 kg mass is attached to a light spring with a force constant of 34.9 N/m and set into oscillation on a horizontal friction

Whitepunk [10]

(a) 0.456 m/s

The maximum speed of the oscillating mass can be found by using the conservation of energy. In fact:

- At the point of maximum displacement, the mechanical energy of the system is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$ (1)

where

k = 34.9 N/m is the spring constant

A = 5.0 cm = 0.05 m is the amplitude of the oscillation

- At the point of equilibrium, the displacement is zero, so all the mechanical energy of the system is just kinetic energy:

$E=K=\frac{1}{2}mv_{max}^2$ (2)

where

m = 0.42 kg is the mass

vmax is the maximum speed, which is maximum when the mass passes the equilibrium position

Since the mechanical energy is conserved, we can write (1) = (2):

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{(34.9 N/m)(0.05 m)^2}{0.42 kg}}=0.456 m/s$

(b) 0.437 m/s

When the spring is compressed by x = 1.5 cm = 0.015 m, the equation for the conservation of energy becomes:

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (3)

where the total mechanical energy can be calculated at the point where the displacement is maximum (x = A = 0.05 m):

$E=\frac{1}{2}kA^2=\frac{1}{2}(34.9 N/m)(0.05 m)^2=0.044 J$

So, solving (3) for v, we find the speed when x=1.5 cm:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s$

(c) 0.437 m/s

This part of the problem is exactly identical to part b), since the displacement of the mass is still

x = 1.5 cm = 0.015 m

So, the speed when this is the displacement is

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s$

(d) 4.4 cm

In this case, we have that the speed of the mass is 1/2 of the maximum value, so:

$v=\frac{v_{max}}{2}=\frac{0.456 m/s}{2}=0.228 m/s$

And by using the conservation of energy again, we can find the corresponding value of the displacement x:

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2\\x=\sqrt{\frac{2E-mv^2}{k}}=\sqrt{\frac{2(0.044 J)-(0.42 kg)(0.228 m/s)^2}{34.9 N/m}}=0.044 m=4.4 cm$

4 0

3 years ago

What average net force is required to accelerate a car with a mass of 1200 kg from rest to 27.0 m/s2 in 10.0 s ?

pogonyaev

Use formula F = MA
Acceleration= (27 m/s2)/10s = 2.7 m/s/s
F = 1200kg * 2.7 m/s/s = 3240 N

5 0

3 years ago