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3 years ago

PLS HELP WILL MARK BRAINLIEST IF RIGHT NEED IMMEDIATELY PLEASEEEEE

Physics

2 answers:

andrew11 [14]3 years ago

5 0

Answer:assisting and caring for others

making decisions and solving problems

getting information

Explanation:

I'm pretty sure about the answer. Thank you

timofeeve [1]3 years ago

3 0

Answer:

Explanation:

Prescribe medications.

Prescribe treatments or therapies.

Treat patients using psychological therapies.

Collect medical information from patients, family members, or other medical professionals.

Record patient medical histories.

Develop medical treatment plans

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I MARK BRAINLIEST, PLEASE ANSWER ASAP!!! PHYSICS 20 POINTS!!

irina [24]

The first one is C. Hope this helps!! If you need anymore help just message me and I will try to get back to you quickly and help in any way i can!

5 0

3 years ago

Read 2 more answers

A 0.90-kg falcon is diving at 28.0 m/s at a downward angle of 35° . It catches a 0.325-kg pigeon from behind in midair. What is

pashok25 [27]

Answer:

22.11 m / s

Explanation:

The falcon catches the prey from behind means both are flying in the same direction ( suppose towards the left )

initial velocity of falcon = 28 cos 35 i - 28 sin 35 j

( falcon was flying in south east direction making 35 degree from the east )

momentum = .9 ( 28 cos 35 i - 28 sin 35 j )

= 20.64 i - 14.45 j

initial velocity of pigeon

= 7 i

initial momentum = .325 x 7i

= 2.275 i

If final velocity of composite mass of falcon and pigeon be V

Applying law of conservation of momentum

( .9 + .325) V = 20.64 i - 14.45 j +2.275 i

V = ( 22.915 i - 14.45 j ) / 1.225

= 18.70 i - 11.8 j

magnitude of V

= √ [ (18.7 )² + ( 11.8 )²]

= 22.11 m / s

6 0

4 years ago

he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee

Crazy boy [7]

Answer:

$F = -1.07 *10^{-8} \ N$

$F_r = 1.07 *10^{-8} \ N$

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

$F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}$

Here k is known as the proportionality constant with value $k = 2.31 * 10^ {-28} J \cdot m$

substituting -2 for $Q_{O}$ i.e the charge on oxygen , +1 for $Q_{Li}$ i.e the charge on Lithium and $[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}$ for r

$F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }$

$F = -1.07 *10^{-8} \ N$

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

$F_r = 1.07 *10^{-8} \ N$

6 0

4 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J

Lyrx [107]

Explanation:

Given that,

Wavelength of the light, $\lambda=4170\ A=4170\times 10^{-10}\ m$

Work function of sodium, $W_o=4.41\times 10^{-19}\ J$

The kinetic energy of the ejected electron in terms of work function is given by :

$KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J$

The formula of kinetic energy is given by :

$KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s$

Hence, this is the required solution.

7 0

3 years ago