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DanielleElmas [232]
3 years ago
8

A 11 kg mass on a frictionless inclined surface is connected to a 2.1 kg mass. The pulley is massless and frictionless, and the

connecting string is massless and does not stretch. The 2.1 kg mass is acted upon by an upward force of 6.6 N, and thus has a downward acceleration of only 3.6 m/s 2 . The acceleration of gravity is 9.8 m/s 2 . What is the tension in the connecting string?
Physics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

6.4 N

Explanation:

Draw free body diagrams for each mass.

For the mass on the inclined surface, the sum of forces parallel to the incline are:

∑F = ma

T - Mg sin θ = Ma

For the hanging mass, the sum of the forces in the y direction are:

∑F = ma

T - mg + F = m(-a)

We're given that g = 9.8 m/s², M = 11 kg, m = 2.1 kg, F = 6.6 N, and a = 3.6 m/s².

From the second equation, we have everything we need to find the tension force.

T - (2.1)(9.8) + 6.6 = (2.1)(-3.6)

T = 6.42 N

Rounded to 2 sig figs, the tension is 6.4 N.

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alexgriva [62]

Answer:

W =23807.68 N

Explanation:

given,

surface area of wing = 19.4 m²

speed over top wing = 67 m/s

speed under wing = 51 m/s

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weight of plane

From Bernoulli's principle

P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho_2^2

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so if we call 1 the lower surface and 2 the upper surface,

we find the pressure differential, P₁ -P₂

\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)

\Delta P =\dfrac{1}{2}\times 1.3 \times (67^2-51^2)

\Delta P =1227.2\ N/m^2

then the force acting on the plane is

F=P A

F=1227.2 x 19.4

F =23807.68 N

weight of the plane

W =23807.68 N

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Ion-dipole forces are the strongest of the intermolecular forces. ... However, hydrogen bonds are still not as strong as ion-dipole interactions. In order from strongest to weakest, the intermolecular forces given in the answer choices are: ion-dipole, hydrogen bonding, dipole-dipole, and Van der Waals forces.

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