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3 years ago

Pleaseeee hurry!!!

Physics

1 answer:

atroni [7]3 years ago

7 0

Answer: C

Explanation: this is called an isotope which has equal number of protons but different numbers of neutrons in the nucleus hence different atomic mass but no difference in chemical properties.

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The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m

Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

$\theta=\frac{s}{r}$

For the Moon

$\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad$

The angle subtended by the Moon is 0.00903 rad

For the Sun

$\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad$

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

$\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}$

The area ratio is $6.268\times 10^{-6}$

3 0

3 years ago

The bodies in this universe attract one another name the scientist who propounded this statement

Colt1911 [192]

<h2><em>Answer: According to wikipedia Sir Isaac Newton has told this statement.</em></h2><h2><em>hope its helps you.</em></h2><h2><em>have a great day.</em></h2><h2><em> keep smiling be happy stay safe. </em></h2><h2><em /></h2>

<em />

7 0

2 years ago

2. physical quantities which do not depend on any physical quantities for their measurements are known as

seropon [69]

Answer:

Fundamental quantities

5 0

2 years ago

What is a benefit of increased physical fitness

8_murik_8 [283]

The benefit's of increased physical fitness is better health, more pleasing appearance to some, stronger muscles and the reduction of fat.

5 0

3 years ago

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

<span>On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

3 years ago