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My name is Ann [436]
3 years ago
13

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

cm2 in 3 days, how long will it take for the leaf's area to increase to 5 cm2
Physics
1 answer:
marusya05 [52]3 years ago
3 0

Answer: 9 days

Explanation:

  • Step 1

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

where <em>A</em> is initial area of the leaf, <em>A1</em> is the final area of the leaf and<em> t</em> 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 = 2cm^{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 5cm^{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.

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Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi
jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

n_{e} = number of electrons = 6.25 x 10⁶

q_{e} = charge on electron = - 1.6 x 10⁻¹⁹ C

n_{p} = number of protons = 7.75 x 10⁶

q_{p} = charge on proton =  1.6 x 10⁻¹⁹ C

Net charge is given as

Q = n_{e} q_{e} + n_{p} q_{p}

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

6 0
3 years ago
A projectile is fired vertically upwards and reaches a height of 78.4 m. Find the velocity of projection and the time it takes t
Musya8 [376]

Answer:

1.) U = 39.2 m/s

2.) t = 4s

Explanation: Given that the

height H = 78.4m

The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2

Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0

Velocity of projections can be achieved by using the formula

V^2 = U^2 - 2gH

g will be negative as the object is moving against the gravity

0 = U^2 - 2 × 9.8 × 78.4

U^2 = 1536.64

U = sqrt( 1536.64 )

U = 39.2 m/s

The time it takes to reach its highest point can be calculated by using the formula;

V = U - gt

Where V = 0

Substitute U and t into the formula

0 = 39.2 - 9.8 × t

9.8t = 39.2

t = 39.2/9.8

t = 4 seconds.

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Help wit these questions someone.
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In series circuit, Req = R₁ + R₂ + R₃ + ···

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<h3>Q7.</h3>

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\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}

R₂ + 2 = 12

R₂ = 10Ω

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\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}

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Answer:

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