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3 years ago

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Ira is concerned about his posture. He tests it by seeing how long he can hold good posture and finds that he feels the most dis

comfort in his back. What does that suggest that he needs to do?
A.
Keep a neutral spine.
B.
Place his hands on his hips.
C.
Arc his back when he walks.
D.
Tuck his chin close to his chest.

1 answer:

larisa [96]3 years ago

6 0

Answer: C

Explanation:

JUST TO SIMPLE

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Some help please ,.......

serious [3.7K]

It's c bro, .2x5 equals 1, which accounts for the 1m/s accelerations.

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3 years ago

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A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr

Ipatiy [6.2K]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = $\frac{9\times 10^9\times3\times10^{-9}}{d^2}$

Field due to charge = 4 X 10⁻⁹ C

$E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$

These two fields will be equal and opposite to make net field zero

$\frac{9\times 10^9\times3\times10^{-9}}{d^2}$ = $[tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}$ [/tex]

$\frac{3}{d^2} =\frac{4}{(2-d)^2}$

$\frac{2-d}{d} =\frac{2}{1.732}$

d = 0.928

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3 years ago

Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

$\lambda = 656.46 \ nm$

$\Delta \lambda = 0.04$

$c=3\times 10^8$

As we know,

⇒ $\frac{\Delta \lambda}{\lambda} =\frac{v}{c}$

On substituting the values, we get

⇒ $\frac{656.46}{0.04} =\frac{v}{3\times 10^8}$

⇒ $v=\frac{656.46}{0.04} (3\times 10^8)$

⇒ $=16411.5\times 3\times 10^8$

⇒ $=18280 \ m/s$

or,

⇒ $=18 \ km/s$

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3 years ago

The simplest form of a substance

likoan [24]

Answer :

C) Atom

Actual answer should be element but depends on the way you interpret the question and the options given to answer it

Explanation :

Atom is the most basic unit of any substance and is what molecules are made of.

Hope it helps if it does let me know by thanking

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3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago