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2 years ago

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A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a

nd the other is at one end. What can be said about the forces exerted by the supports?

2 answers:

JulijaS [17]2 years ago

8 0

All the weight of the wooden board is bear by the support located at the centre of the rod, and the other support which is located at the end, will have no reaction force, or 0 reaction force.

Therefore the reaction at the centre support is equal to the weight of the board, while the support at the end has 0 reaction force.

gogolik [260]2 years ago

4 0

Answer:

The support at the middle of the board exerts an upward force that is equal to the weight of the board mg ( 24x9.81)N which is 235.44N.

The support at the other end exerts no force on the board since the board is already balanced by the middle support.

Explanation:

Detailed explanation and calculation is shown in the image below

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A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

Mice21 [21]

Answer:

$v = 3369.2 m/s$

Explanation:

As we know that Beacon is rotating with angular speed

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

now we know that

$v = r \omega$

here we will have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

so we have

$v = 3218(1.047)$

$v = 3369.2 m/s$

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2 years ago

How does energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid?

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Molecules in a gas move faster than in a liquid.

hope it helps.

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1 year ago

Read 2 more answers

A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity

Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches = 0.0508 m

n = 1750 rpm

$H_{nom}=2hp=1491.4W$

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

$w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m$

$V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s$

$F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N$

$(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N$

$T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm$

$F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N$

$F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N$

b)

$H_a=1491*1.25=1863.75W$

$n_f_s=\frac{H_a}{H_{nom}K_S }=1$

dip = $\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm$

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1 year ago

7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in

viva [34]

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

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2 years ago

A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,

Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

$d_x = 10 + 2= 12 blocks$

Total displacement towards South

$d_y = 5 block$

now to find the net displacement we can use vector addition

$d = \sqrt{d_x^2 + d_y^2}$

$d = \sqrt{12^2 + 5^2}$

$d = 13 blocks$

<em>so magnitude of net displacement will be equal to 13 blocks</em>

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1 year ago