Hartman value profile

How quickly a leaf grows is proportional how big [ie the surface area] the leaf is. If the area of the leaf grows from 2cm2 to 3

marusya05 [52]

Answer: 9 days

Explanation:

Step 1

Let the rate of Leaf growth r be defined as, $\frac{Increase in area}{time taken}$ = $\frac{A1 - A}{t}$

where A is initial area of the leaf, A1 is the final area of the leaf and t is the time taken for the increase in Area.

Express the proportional relationship in equation.

Given that rate of leaf growth, r is proportional to the surface area of the leaf A. we have r ∝ A.

r = kA, where k is the rate constant.

therefore, k = $\frac{r}{A}$

when A = 2 $cm^{2}$ , A1 = 3

so k = $\frac{\frac{3 - 2}{3}}{2}$

= $\frac{1}{3}$ ÷ 2

= 0.33 ÷ 2

k = 0.167

After calculating the rate constant k, we then find the time t when A1 is 5 $cm^{2}$

we have r = k × A1 = $\frac{A1 - A}{t}$

so, 0.167 × 2 = $\frac{5 - 2}{t}$

0.33 = $\frac{3}{t}$ .

t = 3/0.33

Therefore, t = 9 days.

3 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 15.0 ∘ west of north,

Darina [25.2K]

Answer:

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

Explanation:

The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.

The angle between the 2 forces and displacement is ∅ = 15°.

First we have to calculate the work done by the individual force and then we can calculate the total work.

The work done on a particle by a constant force F during a straight line displacement s is given by following formula:

W = F*s

W = F*s*cos∅

With ∅ = the angles between F and s

The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N

The total work done can be calculated as followed:

Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2

Wtotal = 2Fs*cos∅

Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°

Wtotal = 3.44*10^9 Nm = 3.44 *10^9 J

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

5 0

2 years ago

A message is sent from the Galileo spacecraft orbiting Jupiter to earth at a distance of 928,000,000km. If it took the signal 51

expeople1 [14]

The answer would approximately be 299,741.60

6 0

3 years ago

Read 2 more answers

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

2 years ago

An object with kinetic energy k explodes into two pieces, each of which moves with twice the speed of the original object.

zlopas [31]

Assuming that the momenta of the two pieces are equal: when they have equal velocities, then the masses of the two pieces are also equal. Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90Â° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities. This is only possible, when the angle of their velocity with the initial direction is 60Â°. Because, cos (60Â°) = 1/2 = v/(2v).

6 0

3 years ago