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2 years ago

What is the formula for calculating the efficiency of a heat engine?

Physics

2 answers:

Anton [14]2 years ago

7 0

Answer:

Efficiency = StartFraction T Subscript h Baseline minus T Subscript C Baseline over T Subscript h Baseline EndFraction times 100. Efficiency equals T Subscript c Baseline minus T Subscript h Baseline over T Subscript h Baseline all times 100.

Neko [114]2 years ago

3 0

Answer:

Explanation:

BCUZ

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In cats, short hair is dominant and long hair is recessive. If a cat has the genotype of hh, what type of hair will it have?

raketka [301]

Answer:

long hair

Explanation:

long hair = hh

short hair= HH

and medium= Hh

8 0

2 years ago

Can I get a direct answer please??

OleMash [197]

Get a direct answer of what???

7 0

3 years ago

A paleontologist estimates that when a particular rock formed, it contained 12 mg of the radioactive isotope potassium-40, which

leva [86]

Answer:

$t = 2.52 billion \:years$

Explanation:

As we know by radioactivity law

$N = N_o e^{-\lambda t}$

so here we will have

$N = 3 mg$

$N_o = 12 mg$

now we will have

$3 = 12 e^{-\lambda t}$

$\lambda t = ln 4$

now we also know that

$\lambda = \frac{ln2}{1.26 \times 10^6 yrs}$

$t = 1.26\times 10^6\times \frac{ln4}{ln2}$

$t = 2.52 billion \:years$

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2 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

5 0

2 years ago

A negative charge, q1, of 6 µC is 0. 002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the e

prohojiy [21]

Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

<h3>What is electrical force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is

q₁ is the negative charge = 6 µC=6×10⁻⁶ C

q₂ is the positive charge = 3 µC=3×10⁻⁶ C

r is the distance between the charges=0.002 m

$F_E$ is the electric force =?

The value of electric force will be;

$\rm F_E= \frac{Kq_1q_2}{r^2} \\\\ F_E= \frac{9\times 10^9\times 6\times 10^{-6}\times3\times10^{-6}}{(0.002)^2}\\\\ \rm F_E=4.05\times10^4\;N$

Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

To learn more about the electrical force refer to the link;

brainly.com/question/1076352

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1 year ago