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Lapatulllka [165]

2 years ago

9

The atomic number of magnesium is 12. This means that its nucleus must contain

2 answers:

valentinak56 [21]2 years ago

5 0

D took test c:

*have a great day/evening c: *

lana [24]2 years ago

4 0

<span>The correct answer is D) 12 protons and no electrons. This is correct because the number of protons in the nucleus of the atom is 12. The number of protons in the nucleus represents the atomic number. Therefore if there are 12 protons in the nucleus, the atomic number is 12.</span>

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What alleles are always both expressed

lawyer [7]

An organism which has two different alleles of the gene is called heterozygous. Phenotypes (the expressed characteristics) associated with a certain allele can sometimes be dominant or recessive, but often they are neither.

5 0

2 years ago

Read 2 more answers

A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c

den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, $L_1=L-1$

New time period of the pendulum is, $T_1=2.81\ s$

We know that, the time period of a simple pendulum of length 'L' is given as:

$T=2\pi\sqrt{\frac{L}{g}}$ -------------- (1)

So, for the new length, the time period is given as:

$T_1=2\pi\sqrt{\frac{L_1}{g}}$ ------------ (2)

Squaring both the equations and then dividing them, we get:

$\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1$

Now, plug in the given values and calculate 'L'. This gives,

$L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m$

Therefore, the original length of the simple pendulum is 2.97 m

4 0

2 years ago

The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli

Elenna [48]

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

<u>C. 10⁻³ rads</u>

4 0

2 years ago

To live deep ocean what condition do a fish need to be adapted for

tekilochka [14]

High pressure
Low oxygen
Darkness

7 0

2 years ago

Find the solution set of x + y = 5, 2 x - y = 7

STALIN [3.7K]

X + y = 5
x = 5 - y

2x - y = 7
2(5-y) - y = 7
10 - 2y - y = 7
10 - 3y = 7
10 - 7 = 3y
3 = 3y
y = 1

x + y = 5
x + 1 = 5
x = 5 - 1
x = 4

4 0

2 years ago