The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.
Define work. Explain the rate of doing work.
Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.
Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.
Solution Explained:
Given,
Weight = 1000N and distance = 5m
A/Q, the work here is done in lifting then
Work = (weight) × (distance moved)
= 1000 X 5
= 5000Nm or 5000J = 5kJ
Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.
To learn more about work, use the link given
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Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt
- Integrate and evaluate the on the interval:
- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:
- Integrate and evaluate the on the interval:
Answer: if most people are driving the way that you are
Explanation:the law of the many
Answer:
1.1451 x 104 (11451.13)mg/m3
Explanation:
1 ppmv is defined as one volume of a contaminant or solid(CO)(mL) in 1 x 106 volume of solvent/water.
1ppmv = 1mL/m3
Concentration in mg/m3 = volume in ppm x molecular weight x pressure(kPa)/( gas constant x temperature(K)
Molecular weight of CO = 12 + 16
= 28g/mol
Temperature = 273.15 + 25
= 298.15K
Pressure = 1 x 101.325kPa
= 101.325kPa
Ppmv = 1 x 10-4ppmv
Gas constant, R = 8.3144 L.kPa/mol.K
Concentration in mg/m3 = (1 x 104 * 28 * 101.325)/(8.3144 * 298)
= 1.1451 x 104mg/m3
= 11451.13 mg/m3
Answer:
The level of the service is loss and the density is 34.2248 pc/mi/ln
Explanation:
the solution is attached in the Word file
