According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:

(1)
we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:
- the horizontal component of the force exerted by the rope, which is equal to

- the frictional force, acting in the opposite direction, which is equal to

By applying Newton's law (1), we can calculate the acceleration of the box:

Answer:
Yes
Explanation:
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Answer:
<em>11.06m/s²</em>
Explanation:
According to Newtons second law of motion

Given
Mass m = 17kg
Fm = 208N
theta = 36 degrees
g = 9.8m/s²
a is the acceleration
Substitute
208 - 0.148(17)(9.8)cos 36 = 17a
208 - 24.6568cos36 = 17a
208 - 19.9478 = 17a
188.05 = 17a
a = 188.05/17
a = 11.06m/s²
<em>Hence the the magnitude of the resulting acceleration is 11.06m/s²</em>
Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ
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