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1 year ago

a log splitter puts out 1 100W of power for every 1500 W put into it. the efficiency of the machine is what

Physics

2 answers:

hram777 [196]1 year ago

8 0

Answer:400 is the answer 1500-1100 is 400

Explanation:

Lapatulllka [165]1 year ago

4 0

η=out/in x 100%

η=1100/1500 x 100%

η=73%

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Pls answer it ASAP I need it today

Lapatulllka [165]

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4 0

2 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

3 years ago

On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced

Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

$v =f\lambda$

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

$\mu = \frac{m}{l}$

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

$\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m$

Now, for the tension we use the formula:

$v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T$

T = 0.015 N

(b)

The mass to be hung is:

$T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\$

M = 1.53 x 10⁻³ kg = 1.53 g

4 0

3 years ago

A 0.0450-kg golf ball initially at rest is given a speed of 25.2 m/s when a club strikes. part a part complete if the club and b

Ksenya-84 [330]

We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s

To find force we use formula:
F = m * a

a is acceleration. To find it we use formula:
a = Δv / Δt
a = 25.2 / 0.00195
a = 12923.1 m/s^2

Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N

To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
W = m * g
W = 0.0450 * 9.81
W = 0.44145 N

We can see that weight is much smaller than the applied force so it's influence in negligible.

3 0

3 years ago

HELP ME 20 POINTS!!!Assume that the average volume of an adult human body is one-tenth

Agata [3.3K]

Answer:

,Assume that the average volume of an adult human body is one-tenth

cubic meter (0.10 m) and that there are two billion (2.0 x 109)

adults in the world.

a. What would be the total volume of all the adults in the world?

b. Compute the length of one edge of a cubic container that has a

volume equal to the volume of all the adults in the world.