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kolezko [41]
4 years ago
13

â/8.37 points scalcet8 12.2.037. ask your teacher my notes question part points submissions used a block-and-tackle pulley hoist

is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 350 n. the ropes, fastened at different heights, make angles of 50° and 38° with the horizontal. find the tension in each rope and the magnitude of each tension. (let t2 and t3, represent the tension vectors corresponding to the ropes of length 2 m and 3 m respectively. round all numerical values to two decimal places.)

Physics
1 answer:
AnnyKZ [126]4 years ago
4 0
Refer to the diagram shown below.

The hoist is in static equilibrium supported by tensions in the two ropes.

For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂             (1)

For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350     (2)

Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979

Answer:
T₂ = 225.12 N
T₃ = 275.98 N

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Answer:

C. 72

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In a ideal transformer,

Vs/Vp = Ns/Np ............................................. Equation 1

Where Vp = primary voltage, Vs = secondary voltage, Ns = Secondary turn, Np = primary turn.

Making Ns the subject of the equation,

Ns =(Vs/Vp)Np .......................................... Equation 2

Given: Vs = 24 V, Vp = 115 V, Np = 345.

Substitute into equation 2

Ns = (24/115)345

Ns = 72 turns.

Thus the number of turns in the secondary = 72 turns.

The right option is C. 72

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Answer:

reviewing the opinion of the two students we see that neither is right, since when the kinetic energy increases the potential energy decreases by the same value

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From this expression we see that when an electron moves from the initial point to the final point, the potential energy must decrease, for the total energy to be constant.

When reviewing the opinion of the two students we see that neither is right, since when the kinetic energy increases the potential energy decreases by the same value

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Fine grains of beach sand are assumed to be spheres of radius 64.8 µm. These grains are made of silicon dioxide which has a dens
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Answer:

1) Mass of the grain is 2.9632\times 10^{-9} kg.

2) 0.08901 kg of sand would have surface area equal the surface area of the cube.

Explanation:

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Volume of the sphere =\frac{4}{3}\pi r^3

Volume of the grain of a sand:

V=\frac{4}{3}\times \pi r^3=\frac{4}{3}\times 3.14\times (6.48\times 10^{-5} m)^3

V=1.1397\times 10^{-12} m^3

Density of a grain of sand = d=2600 kg/m^3

Mass of a grain of a sand = M

d=2600 kg/m^3=\frac{M}{1.1397\times 10^{-12} m^3}

M=2.9632\times 10^{-9} kg

Mass of the grain is 2.9632\times 10^{-9} kg.

2) Surface are of sphere: 4\pi r^2

Surface area of a grain:

A=4\times 3.14\times (6.48\times 10^{-5}m)^2

A=5.2766\times 10^{-8} m^2

Length of the cube = a = 0.514 m

Total surface area of cube ,A'= 6a^2

A'=6\times (0.514 m)^2=1.5851 m^2

let the number grains with area equal to total surface area of cube be x.

A'=A\times x

x=\frac{1.5851 m^2}{5.2766\times 10^{-8} m^2}=3.003\times 10^7

Volume of x number of grains :V'

V'=V\times x

V= 1.1397\times 10^{-12} m^3\times 3.003\times 10^7

V'=3.4236\times 10^{-5} m^3

Mass of 3.4236\times 10^{-5} m^3 of sandL:

=3.4236\times 10^{-5} m^3\times 2600 kg/m^3=0.08901 kg

0.08901 kg of sand would have surface area equal the surface area of the cube.

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