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Dmitry_Shevchenko [17]
3 years ago
5

A tissue cross-section viewed on a microscope slide shows a cell of almost rectangular shape with dimensions 6.9 × 10-4 cm by 1.

1 × 10-3 cm. Calculate the cross-sectional area of the cell. Express the area in powers of ten notation and use the appropriate number of significant digits for the number multiplying the power of 10.
Physics
1 answer:
zimovet [89]3 years ago
4 0

Answer:

The  cross-sectional  area is A = 7.59 *10^{-11} \  m^2

Explanation:

From the question we are told that

   The dimension of the cell is    6.9 × 10-4 cm by 1.1 × 10-3 cm = 6.9*10^{-6} \ m \ by\  1.1*10^{-5} \  m

   Generally the cross-sectional area is mathematically represented  as

    A  =  6.9*10^{-6}   *  1.1*10^{-5}

=>   A = 7.59 *10^{-11} \  m^2

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A ping pong ball has a mass of about 2.45 grams. Suppose that Forrest Gump hits the ball across the table with a speed of about
KonstantinChe [14]

Answer:

0.0196 Joules

Explanation:

Kinetic energy is half the product of mass and velocity squared.

K.E=0.5 m*v^2; The m is the mass in kilograms and v is the velocity in meters per second.

We will convert grams to kilograms:

2.45 grams = 0.00245 kilograms.

Calculating kinetic energy:

mass:0.00245 kg

velocity:4 m/s

K.E = 0.5 *0.00245*4^2 = 0.0196 Joules.

4 0
3 years ago
at a 200 meter swimming competition bill got a personal record of 45 secomda and time was 5 secomds faster then bill at the comp
steposvetlana [31]

Answer:

Average speed of the swimmer is 4.44 m/s

Explanation:

As we know that the speed is the ratio of distance covered over time interval

so here we can say

v = \frac{d}{t}

so we know that

d = 200 m

time taken to complete the trip is

t = 45 s

now we have

v = \frac{200}{45}

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4 0
4 years ago
Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardr
Margaret [11]

Answer:

a. d₁/d₂ = 1.09 b. 0.054 mW

Explanation:

a. What is the ratio of the diameter of the first student's eardrum to that of the second student?

We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²

So, I ∝ I/d²

I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.

Given that I₂ = 1.18I₁

I₂/I₁ = 1.18

Since I₁/I₂ = d₂²/d₁²

√(I₁/I₂) = d₂/d₁

d₁/d₂ = √(I₂/I₁)

d₁/d₂ = √1.18

d₁/d₂ = 1.09

So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09

b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?

We know intensity, I = P/A where P = acoustic power and A = area = πd²/4

Now, P = IA

= I₂A₂

= I₂πd₂²/4

= 1.18I₁πd₂²/4

Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m

So, P = 1.18I₁πd₂²/4

= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4

= 0.691244π × 10⁻⁴ W/4 =

2.172 × 10⁻⁴ W/4

= 0.543 × 10⁻⁴ W

= 0.0543 × 10⁻³ W

= 0.0543 mW

≅ 0.054 mW

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3 years ago
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MgCl2 is magnesium chloride.
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8 0
4 years ago
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Answer:

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