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1 year ago

I

Physics

1 answer:

Anna [14]1 year ago

3 0

Answer:

Here, m=10 kg

The resultant force acting on the body is

F=(98N)2+(6N)2=10N

Let the resultant force F makes an angle θ w.r.t. 8N force.

From figure, tanθ=8N6N=43

The resultant acceleration of the body is

a=mF=10kg10N=1ms−2

The resulatnt acceleration is along the direction of the resulatnt force.

Hence, the resultant acceleration of the body is 1 ms−2 at an angle of tan−1(43) w.r.t. 8N force.

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A 0.22-caliber handgun fires a 27-g bullet at a velocity of 755 m/s. is the wave nature of matter significant for bullets?

Neporo4naja [7]

<span>All matter can exhibit wave-like behavior, but the question asks if it is significant. A bullet is a very, very large object compared to the light wave-particles where the wave nature is typically observed. So for an object of this magnitude, it is insignificant.</span>

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3 years ago

If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti

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3 years ago

The average velocity of an object over 6.0 seconds interval is 2 m/s what is the total distance traveled and M by the object doi

lina2011 [118]

Answer:

<h2>The answer is 12 m</h2>

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2 years ago

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

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ira [324]

Can't really plot a graph here for question 1.

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