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2 years ago

A machine has an efficiency of 80% how much work must be done on the machine so to make it do 50000 J of outfit work

1 answer:

6 0

With efficiency of 80%, the machine outputs 50000 J. The input to the machine should represent 100%.

If
80% --- 50000 J
100% --- ?? J

Solution:
Input work = (100/80)*500000 J = 62500 J

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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

stellarik [79]

Answer:

0.243 m/s

Explanation:

From law of conservation of motion,

mu+m'u' = V(m+m')................. Equation 1

Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.

make V the subject of the equation

V = (mu+m'u')/(m+m')................. Equation 2

Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s

Substitute into equation 2

V = (260000×0.32+52500×(-0.14))/(260000+52500)

V = (83200-7350)/312500

V = 75850/312500

V = 0.243 m/s

8 0

2 years ago

1. An object has a mass of 10g and a volume of 50mL. What is the density?

Alona [7]

1. 0.2 g/mL

The relationship between mass, density and volume of an object is

$d=\frac{m}{V}$

where

d is the density

m is the mass

V is the volume

For the object in this problem, we have

m = 10 g

V = 50 mL

Substituting into the equation,

$d=\frac{10}{50}=0.2 g/mL$

2. 10 mL

In this exercise we know:

- The density of the object: d = 2 g/mL

- The mass of the object: m = 20 g

Therefore, we can re-arrange the previous equation to find the volume:

$V=\frac{m}{d}$

And substituting values into the equation, we find

$V=\frac{20}{2}=10 mL$

7 0

2 years ago

1. Consider the point exactly halfway between the two wires. Can you adjust the current so that, with current passing through ea

chubhunter [2.5K]

Answer:

hello your question is incomplete attached below is missing part of the question

answer:

1 ) Magnetic field due to long current carrying wire : $B = \frac{U_{0} I}{2\pi d}$

Therefore the net magnetic field due the both wires ; B = B $_{1}$ + B $_{2}$ . when we adjust the current I $_{1}$ = I $_{2}$ then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

$F = \frac{U_{0}I_{1}I_{2} }{2\pi d }$

Explanation:

1 ) Magnetic field due to long current carrying wire : $B = \frac{U_{0} I}{2\pi d}$

Therefore the net magnetic field due the both wires ; B = B $_{1}$ + B $_{2}$ . when we adjust the current I $_{1}$ = I $_{2}$ then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

$F = \frac{U_{0}I_{1}I_{2} }{2\pi d }$