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3 years ago

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The orbit of a certain a satellite has a semimajor axis of 4.0 x 107 m and an eccentricity of 0.15. Its perigee (minimum distanc

e) and apogee (maximum distance) are respectively

1 answer:

Mashutka [201]3 years ago

5 0

Answer:

100KM and 1kkm

Explanation:

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Radioactive iodine (radioiodine) is given to patients with some forms of thyroid cancer in order to destroy the cancer cells. 13

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Explanation:

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3 years ago

A metal cylinder measures 3.0 cm in diameter, is 12.0 cm high, and has a mass of 750 g. The cylinder hangs from a string attache

kvasek [131]

Answer with Explanation:

We are given that

Diameter of cylinder, $d_1=$ 3 cm

Radius, $r_1=\frac{d_1}{2}=\frac{3}{2}=1.5cm=1.5\times 10^{-2} m$

1m=100 cm

Height,h=12 cm= $\frac{12}{100}=0.12 m$

Mass,m=750 g= $\frac{750}{1000}=0.75kg$

1kg=1000 g

Diameter of beaker, $d_2=10 cm$

Radius of beaker, $r_2=\frac{d_2}{2}=\frac{10}{2}=5cm$

A. $\pi r^2_2h_2=\pi r^2_1 h_1$

$h_2=\frac{r^2_1}{r^2_2}h$

$h_2=\frac{(1.5)^2}{(5)^2}\times 12=1.08 cm$

Hence, the height of the water in the beaker rises when the cylinder is submerged=1.08 cm

B.Weight read on the scale after the cylinder is submerged

$w=mg-\rho A_1h_1g$

Where $g=9.8 m/s^2$

$A_1=\pi r^2_1$

Density of water= $\rho=1000kg/m^3$

$w=0.75\times 9.8-1000\times \pi(1.5\times 10^{-2})^2\times 0.12\times 9.8$

$w=6.52 N$

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3 years ago

Two protons, starting several meters apart, are aimed directly at each other with speeds of 2.00 * 105 m>s, measured relative

victus00 [196]

Explanation:

Since, the two protons are at the initial state of point a and point b. Hence, total mechanical energy at point a and point b is as follows.

$U_{a} + K_{a} = U_{b} + K_{b}$

or, $K_{a} = U_{b}$

where, $K_{a}$ = kinetic energy of two protons

So, $2(\frac{1}{2}mV^{2}_{a}) = \frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r_{b}}$

$r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}$

Putting the given values into the above formula we will calculate the value of $r_{b}$ as follows.

$r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}$

= $(9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}^{2} C}{1.67 \times 10^{-27} kg \times 2 \times 10^{5}}$

= $3.45 \times 10^{-12} m$

Now, we will calculate the maximum electric field as follows.

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r^{2}_{b}}$

= $9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}C)^{2}}{3.45 \times 10^{-12}}$

= $0.194 \times 10^{-4} N$

Therefore, we can conclude that the maximum electric force that these protons will exert on each other is $0.194 \times 10^{-4} N$ .

7 0

3 years ago

Suppose a ray of light traveling in a material with an index of refraction na reaches an interface with a material having an ind

ivolga24 [154]

Answer: Option (b) is the correct answer.

Explanation:

It is known that when a ray of light tends to travel from a denser to rarer medium then there occurs total internal reflection.

For a denser medium the refractive index is greater than that of a rarer medium. This means that for the given situation refractive index of medium $n_{a}$ is greater than medium $n_{b}$ .

this also means that incident angle must be greater than the critical angle of the medium.

Thus, we can conclude that the statement $n_{a}$ > $n_{b}$ must be true for total internal reflection to occur.

5 0

3 years ago