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2 years ago

One mole of an unknown sample contains 120 g of carbon and 30.3 g of

Physics

1 answer:

barxatty [35]2 years ago

5 0

atomic mass of carbon is C = 12 g/mol

atomic mass of hydrogen is H = 1 g/mol

now number of carbon atoms

$N_c = \frac{120}{12} = 10$

Similarly the number of hydrogen atoms are

$N_H = \frac{30.3}{1} = 30$

so Carbon atom and hydrogen atom must be in ratio of 10:30

so the ratio is 1:3

now the empirical formula is always in simplest form

$CH_3$

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3 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

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(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

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$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

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$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time t is

$v_{y,f}=v_{y,i}-gt$

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

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which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

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$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is

$y=v_{y,i}t-\dfrac12gt^2$

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3 years ago

A certain spring stores 10.0J of potential energy when it isstretched by 2.00cm from its equilibrium position.How much potential

qaws [65]

Answer:

Answered

Explanation:

x= 0.02 m

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10= 0.5k(0.02)^2

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K= 50.0 N/m

Now

E'_p= 0.5kx'^2

E'_p= 0.5×50×(0.04)^2

E'_p=40 J

b) potential energy is a scalar quantity and it only depends magnitude and not direction so it will remain same in compression and expansion both

c) 20 J = 0.5×50,000×x^2

solving

x= 0.028 m

d) k is 50.0 N/m from above calculation

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3 years ago

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natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

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