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Crank
4 years ago
11

Bales of hay of 40 lbf move up a conveyor set at 30° angle to the ground. If the hay bales are moving at 1.5 ft/sec, determine t

he power required to move each bale, neglecting any wind or air resistance.
Physics
1 answer:
jekas [21]4 years ago
5 0

Answer:

The power required to move each bale is 30 lbf.ft/sec.

Explanation:

F= 40lbf * sin (30º)

F=20lbf

P= F*v

P=20 lbf * 1.5 ft/sec

P= 30 lbf.ft/sec

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A uniform rod is attached to a wall by a hinge at its base. The rod has a mass of 6.5 kg, a length of 1.3 m, is at an angle of 1
liubo4ka [24]

Answer:

tension wire 104 N, horizontal force hinge 104 N, vertical force hinge 63.7 N

Explanation:

The system described is in static equilibrium under the action of several forces, which are shown in the attached diagram, where T is the tension of the wire, W is the weight of the bar, Fx is the horizontal reaction of the wall and Fy It is the vertical reaction of the wall.

The directions of the forces are indicated by the arrows and are marked intuitively, but when solving the problem if one gives a negative value this indicates that the direction is wrong, but does not alter the results of the problem

For the resolution we use Newton's second law, both in translational and rotational equilibrium, if necessary.

We must establish a reference system to assign the positive meaning, we place it with the origin in the hinge, and the positive directions to the right and up. The location of the coordinate system allows us to eliminate the reaction of the hinge by having zero distance to origin. We write the equilibrium equations for each axis

     ∑ Fx = 0

     Fx -T =0

     ∑ Fy =0

     Fy -W =O      W = m g

     ∑τ =0

      -T dy + W dx =0

For rotational equilibrium we take the positive direction as counterclockwise rotation and distance is the perpendicular distance of the force to the axis of the coordinate system

     

       dy = L sin 17

      dx = (L/2) cos 17

Where L is the length of the bar and the weight is applied at the center of it, we write and simplify the equation

    -T L sin 17 + mg (L / 2) cos 17 = 0

   - T sin 17 + mg/2  Cos 17 =0

We write the three equations together

     Fx- T = 0

     Fy -W = 0

     -T sin 17 + (m g / 2) cos 17 = 0

a) With the third equation we can find the wire tension

     T = m g / 2 cos 17 / sin17

     T = 6.5  9.8/2 cotan 17

     T = 104 N

b) We use the first equation to find Fx

       Fx- T =0

       Fx = T = 104 N

c) We use the second equation to find Fy

 

       Fy = W = m g

       Fy = 6.5 9.8

       Fy = 63.7 N

4 0
3 years ago
A granite monument has a volume of 25,365.4 cm3. The density of granite is 2.7 g/cm3. Use this information to calculate the mass
LiRa [457]
As density = mass/volume

So

Mass = density *volume

Mass = 25,365.4 * 2.7 = 68,486.58 g

<span>Mass of the granite monument to the nearest tenth = 68,485.6 g</span>

5 0
3 years ago
How does the kinetic energy of the particles in a hot cup of coffee compare to the kinetic energy of the particles in ice cream?
trasher [3.6K]

Answer:

it has more molecules than the burning match, which equals MORE total kinetic energy.

Explanation:

4 0
4 years ago
Read 2 more answers
A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some sola
irakobra [83]
C I’m pretty sure!!!!!
4 0
3 years ago
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi
trapecia [35]

Answer:

a)  F_p=882N

b)  P=4410W

c)  V_p'=24135 ,n=15.2\%

Explanation:

From the question we are told that:

Mass M=1500kg

Velocity v=4.9m/s

Coefficient of Rolling Friction \mu=0.06

a)

Generally the equation for The Propulsion Force is mathematically given by

 F_p=\mu*mg

 F_p=0.06*1500*9.81

 F_p=882N

b)

Therefore Power Required at

 V_p=5.0m/s

 P=F_p*V_p

 P=882*5

 P=4410W

c)

 V_p' =15mpg

 V_p'=15*\frac{1609}

 V_p'=24135

Generally the equation for Work-done is mathematically given by

 W=F_p*V_p'

 W=882*15*1609

 W=2.13*10^7

Therefore

Efficiency

 n=\frac{W}{E}*100\%

Since

Energy in one gallon of gas is

 E=1.4*10^8J

Therefore

 n=\frac{2.1*10^7}{1.4*10^8}*100\%

 n=15.2\%

7 0
3 years ago
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