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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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All the compounds are covalent compounds . This means that they are formed by the sharing of pair of electrons.

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A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel
nexus9112 [7]

Answer:

The electric flux is 280\ \rm N.m^2/C

Explanation:

Given:

  • Radius of the disc R=0.50 m
  • Angle made by disk with the horizontal \theta=30^\circ
  • Magnitude of the electric Field E=713.0\ \rm N/C

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

\phi=\int E.dA

where

  • \phi is the total Electric Flux
  • E is the Electric Field
  • dA is the Area through which the electric flux is to be calculated.

Now according to question we have

=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C

Hence the electric flux is calculated.

8 0
3 years ago
In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

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3 years ago
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