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Oksana_A [137]
3 years ago
5

Which statement correctly explains mechanical advantage?

Physics
1 answer:
Shtirlitz [24]3 years ago
4 0
I'm not entirely sure but I would think it's A.
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A commonly used unit of mass in the English system is the pound-mass, abbreviated lbm, where 1 lbm = 0.454 kg. What is the densi
Gemiola [76]

Answer: The density of glycerine will be 0.028lbm/foot^3

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given:

Density of glycerine= 1.26\times 10^3kg/m^3

1 lbm = 0.454 kg

1 kg =\frac{1}{0.454}=2.2lbm

Thus 0.454kg=\frac{2.2}{1}\times 0.454=0.99

Also 1m^3=35.3147foot^3

Putting in the values we get:

Density=\frac{0.99lbm}{35.3147}=0.028lbm/foot^3

Thus density of glycerine will be 0.028lbm/foot^3

3 0
3 years ago
SHOW ADEQUATE WORKINGS IN THIS SECTION
cluponka [151]

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power =  4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort  = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

3 0
3 years ago
The uncertainty Δp sets a lower bound on the average momentum of a particle in the nucleus. If a particle's average momentum wer
kirill [66]

Answer: The minimum kinetic energy Kmin is 1.3 × 10^-13 J

Explanation: Please see the attachments below

8 0
3 years ago
Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In
OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

g

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

\frac{P1}{ρg} +  \frac{V1^{2} }{2g} + Z1 =  \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

\frac{P1}{ρg} + Z1 =  \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3

Z1 = \frac{V3^{2} }{2g} + Z3

V3 = \sqrt{2g(Z1-Z3)}

V3 = \sqrt{2 x 9.8 x (10 - 3)}

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

8 0
3 years ago
find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds
Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}

Final speed of the train, v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}

Displacement of the train, S=234\textrm{ km}=234\times 1000=234000\textrm{ m}

Using Newton's equation of motion,

v - u = at\\a=\frac{v-u}{t}

Now, using Newton's equation of motion for displacement,

v^{2}-u^{2}=2aS

Now, plug in the value of a=\frac{v-u}{t} in the above equation. This gives,

v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}

Now, plug in 234000 m for S, 25 m/s for v and 20 m/s for u. Solve for t.

t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}

Therefore, the time taken by the train is 10400 s.

3 0
3 years ago
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