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Marrrta [24]
4 years ago
8

Which of the following changes would make a heat engine less efficient

Physics
1 answer:
DIA [1.3K]4 years ago
8 0
A heat engine would be less efficient due to many factors 
For instance, a heat engine is more efficient when it uses in cold weather because there is a greater temperature difference ( Carnot Efficient )
A heat engine could be less efficient because of friction
Hope it helps I am a beginner
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An amoeba is a one-celled protist. Amoeba contain all the organelles of a typical eukaryotic animal cell. It is a heterotroph an
horrorfan [7]

The correct answer is the amoeba will deploy its pseudopods (cytoplasmic extentions) to capture the prey and phagocyte.

The amoeba most known and probably the most representative of the kind. Large (up to 500 microns), common in stagnant waters, extremely voracious as evidenced by multiple digestive vacuoles.

Amoebae are characterized by a deformable cell body emitting changes of shape, the pseudopods, which allow them to crawl on a support or to capture microscopic prey by phagocytosis.

4 0
4 years ago
If constants aren't given in an experiment, what can be constant in the experiment?
mario62 [17]

Answer:

Anything in an experiment that remains unchanged.

Explanation:

An example could be the temperature of the laboratory room. If there is something that has an effect on an experiment that is not variable, it is a constant. Another constant could be, say, if you were doing calculations with the same amount and kind of fluid throughout the experiment, then that fluid would also be a constant.

6 0
3 years ago
Describe the changes that occur inside a helium balloon as it rises from sea level
zheka24 [161]

Answer:

The balloon expands in the number of particles per cubic centimeter decreases. This happens because as it expands there is a decrease in the density of area. The Dead Sea is a solution that is so dense that you easily float on it.

8 0
3 years ago
Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out ov
Mumz [18]

Answer:

The number of excess electrons are on the negative surface is 4.80\times10^{10}\ electrons

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}

C=0.285\times10^{-10}\ F

We need to calculate the potential

Using formula of potential

V=Ed

Put the value into the formula

V=18000\times1.5\times10^{-2}\ V

V=270\ V

We need to calculate the charge

Using formula of charge

Q=CV

Put the value into the formula

Q=0.285\times10^{-10}\times270

Q=76.95\times10^{-10}\ C

Here, the charge on both the positive and negative  plates

Q=+76.95\times10^{-10}\ C

Q=-76.95\times10^{-10}\ C

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

n=\dfrac{q}{e}

Put the value into the formula

n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}

n=4.80\times10^{10}\ electrons

Hence, The number of excess electrons are on the negative surface is 4.80\times10^{10}\ electrons

8 0
3 years ago
5. If 100N effort is required to lift a load of 300N in a lever of 1m and distance between fulcrums to load is 30 cm then, calcu
pishuonlain [190]

effort distance =?

effort =100N

load =300N

lever =1m

load D =30cm

ma=?

vr=?

efficiency =?

we know that

effort distance =l ×ld

= 300×30

9000

ma= l ×e

= 100×300

= 30000

vr = ld ×ed

= 1×9000

=9000

efficiency =(ma÷vr)/100

(30000÷9000)×100

3.33×100

333

5 0
3 years ago
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