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inn [45]
3 years ago
10

In the presence of ________, hydrogen peroxide will decompose into oxygen and water.

Physics
1 answer:
IrinaK [193]3 years ago
7 0

Answer: Catalase enzyme

Explanation:

The catalase is an enzyme that is commonly found in the living organisms that are exposed to minute or high amount of oxygen or those in which aerobic respiration takes place.

This enzyme catalyzes the breakdown or decomposition of the hydrogen peroxide into oxygen and water. In humans and animals the catalase is present in liver, where it conducts the function of detoxifying the body by decomposing the toxins.

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Light refracts because
NeTakaya

Answer:

it travels at a different refractive index.

Explanation:

for example when light travels into water, it slows down, causing it to continue to travel at a different or another angle or direction. hope this helped!

5 0
4 years ago
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Eukaryotic that can reproduce sexually contain two types of cells
Darina [25.2K]
<span>Only eukaryotes form multicellular organisms consisting of many kinds of tissue made up of different cell types. Eukaryotes can reproduce both asexually through mitosis and sexually through meiosis and gamete fusion. In mitosis, one cell divides to produce two genetically identical cells.</span>
7 0
3 years ago
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject
kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

Radius_{mean} = 60\ mm

Diameter_{mean} = 120\ mm

Load = 200 N

We need to calculate the deflection

Using formula of deflection

\delta=\dfrac{8pD^3n}{Cd^4}

Put the value into the formula

\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}

\delta =34.56\ mm

Hence, The deflection of the spring is 34.56 mm.

4 0
3 years ago
Please answer ASAP!!!
NARA [144]
The third (left hand corner) since the x and y are both negative.
4 0
3 years ago
Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.
Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

\eta = 1-\frac{T_C}{T_H}

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

T_C = 270^{\circ}C+273=543 K

T_H = 450^{\circ}C+273=723 K

Therefore the maximum efficiency is

\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

T_H = 270^{\circ}C+273=543 K is the high-temperature reservoir

T_C = 150^{\circ}C+273=423 K is the low-temperature reservoir

If we apply again the formula of the efficiency

\eta = 1-\frac{T_C}{T_H}

The maximum efficiency of the second engine is

\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221

8 0
3 years ago
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