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3 years ago

10

Which of these is a biotic factor? A) topography B) soil C )air D)bacteria

1 answer:

vitfil [10]3 years ago

7 0

D I might be wroung .....i think this so the answer because biotic are all living things and bacteria is living

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What will be the pressure exerterd by the 0bject if 4000n is acting on an area of 50msqure

Ivahew [28]

<h3>Given, </h3>

Force,F = 4000 N

Area,a = 50 m²

<h3>We know that, </h3>

Pressure = Force/Area

★ Putting the values in the above formula,we get:

$\sf \rightarrow \: pressure = \dfrac{4000}{50}$

$\sf \rightarrow pressure = 80 \: N {m}^{ - 2}$

7 0

3 years ago

the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr

Dmitry [639]

Answer:

The approximate number of decays this represent is $N= 23*10^{10}$

Explanation:

From the question we are told that

The amount of Radiation received by an average american is $I_a = 2.28 \ mSv$

The source of the radiation is $S = 5.49 MeV \ alpha \ particle$

Generally

$1 \ J/kg = 1000 mSv$

Therefore $2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg$

Also $1eV = 1.602 *10^{-19}J$

Therefore $2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg$

An Average american weighs 88.7 kg

The total energy received is mathematically evaluated as

$1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x$

Cross-multiplying and making x the subject

$x = 88.7 * 1.423*10^{16} eV$

$x = 126.2*10^{16}eV$

Therefore the total energy deposited is $x = 126.2*10^{16}eV$

The approximate number of decays this represent is mathematically evaluated as

N = $\frac{x}{S}$

Where n is the approximate number of decay

Substituting values

$N = \frac{126 .2*10^{16}}{5.49*10^6}$

$N= 23*10^{10}$

7 0

3 years ago

Two balls, ball A and ball B, are dropped from the same height onto the same surface. If ball A rebounds to a higher height than

mamaluj [8]

Answer:

its b

Explanation:

3 0

2 years ago

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

7 0

3 years ago

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

$\theta$ = Angle of slide = 20.7°

$\mu$ = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

$W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J$

The work done by the force of gravity is 23.52092 J

8 0

2 years ago