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STALIN [3.7K]
3 years ago
6

While passing through a human cell, an x-ray photon interacts with and inactivates the cell’s master molecule. What is the conse

quence for the cell?
Physics
1 answer:
Yuliya22 [10]3 years ago
3 0

Answer:

Cell Death

Explanation:

Cell death is defined as the biological process which ceases the function of the cell to carry out. This can be caused due to the formation of new cells in place of old cells.

Or it can be cause due to some serious disease or may be caused due to the injury or due to the death of that organism to which these cells belong.

And another case is that when X-ray photon interact with the human cell while it passes through the cell, it   will damage the cell and cease it to function well and a more drastic condition occurs and that cell become dead.  

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Um elétron é lançado entre duas placas eletrizadas como mostra a figura. Sejam v= 6x10^6 m/s, ângulo 45°, E= 2x10^3 N/C, d= 3 cm
Svetlanka [38]
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3 0
3 years ago
A toroidal coil of N turns has a central radius b and a square cross section of side a. Find its self-inductance.
Xelga [282]

Answer:

L = \frac{\mu_0 N^2 (a^2)}{2\pi b}

Explanation:

As we know that magnetic field due to torroid is given as

B = \frac{\mu_0 N i}{2\pi b}

this is approximately constant magnetic field along the axis of the torroid

now the flux linked with one coil of the torroid is given as

\phi = B.A

\phi = \frac{\mu_0 N i}{2\pi b}(a^2)

now total flux of N number of coils is given as

\phi_{total} = \frac{\mu_0 N^2 i(a^2)}{2\pi b}

now we know that self inductance is the property of coil in which flux of the coil will link with the current in the coil

So we know that

L = \frac{\phi}{i}

L = \frac{\mu_0 N^2 (a^2)}{2\pi b}

3 0
3 years ago
1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0
3 years ago
during conversion water to ice,forces of attraction between molecules. OPTIONS:1.)Decreases 2.)Increases 3.)Does not change 4.)C
In-s [12.5K]

because water is loosely packed but when it is cold it becomes closely packed in order to form ice and thus the force attraction between them also increase.

4 0
3 years ago
3) connect two lamps to a power supply in series and current drawn from the power supply is is. connect the same two lamps in pa
notka56 [123]

The current drawn by the series circuit is(R₁ + R₂)/R₁R₂  times the current drawn by the parallel circuit.

Let the resistance of the two lamps are R₁ and R₂.

Then the equivalent resistance in series combination is: R = R₁ + R₂.

And,  the equivalent resistance in parallel combination is:

r = R₁R₂/(R₁ + R₂).

So, if the supply voltage is V,

Then, current drown in series combination; i_s = V/R = V/(R₁ + R₂)

And, current drown in parallel combination; i_p = V/r = V(R₁ + R₂)/R₁R₂

So ,\frac{i_s}{i_p} = [ V/(R₁ + R₂)] /[V(R₁ + R₂)/R₁R₂]

=  (R₁ + R₂)/R₁R₂

Hence, the  ratio of current drawn in series and current drown in parallel is  (R₁ + R₂)/R₁R₂.  So, he current drawn by the series circuit is(R₁ + R₂)/R₁R₂  times the current drawn by the parallel circuit.

Learn more about electric current here:

brainly.com/question/2264542

#SPJ1

4 0
1 year ago
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