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3 years ago

PLSSS HELPP

2 answers:

8 0

The forces shaping the short-term future of man, perhaps to the turn of this century, are apparent, and the events are in train. The shape of the world in the year 2000 and man's place therein will be determined by the manner in which organized humanity confronts several major challenges. If sufficiently successful, and mankind escapes the dark abysses of its own making, then truly will the future belong to man, the only product of biological evolution capable of controlling its own further destiny.

please mark in brain list

photoshop1234 [79]3 years ago

5 0

Answer:

Well, if you look up countries with the highest birth rates, also those countries poverty rates and what class the majority of the people are in then you should have a better understanding. Also, education is not passed through genes it is learned through expirience and being taught. Also being rich vs being poor probably doesn't have that big of an effect either. It is the living conditions.

Explanation:

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3–101 It is estimated that 90 percent of an iceberg’s volume is below the surface, while only 10 percent is visible above the su

Artist 52 [7]

Answer:

The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³

Explanation:

Given that,

Volume of seawater = 90 % of Volume of iceberg

Volume of visible surface = 10 %

Density of seawater = 1025 kg/m³

We need to calculate the density of the iceberg

Using equilibrium condition

$W_{I}=F_{B}$

$\rho_{I}\times V_{I}\times g= \rho_{w}\timesV_{w}\times g$

Put the value into the formula

$\rho_{I}\times V_{I}\times g=\rho_{w}\times0.9V_{I}\times g$

$\rho_{I}=1025\times0.9$

$\rho_{I}=922.5kg/m^3$

We need to calculate the density of the iceberg

Using equilibrium condition

$W_{I}=F_{B}$

$\rho_{I}\times V_{I}\times g= \rho_{w}\timesV_{w}\times g$

Put the value into the formula

$\rho_{I}\times V_{I}\times g=\rho_{w}\times0.1V_{I}\times g$

$\rho_{I}=1025\times0.1$

$\rho_{I}=102.5\ kg/m^3$

Hence, The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³

5 0

3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m2 . A fan is 30 m from the loudspea

marshall27 [118]

The concepts used to solve this problem are sound power, area of ear drum, and inverse square law. First find the intensity $I_2$ using inverse square law for different distances. Then find the area of the ear drum using the diameter and finally find the energy transferred to each ear drum in one second. Intensity is inversely proportional to the square of the distance from the source:

$I \propto \frac{1}{r^2}$

Here,

r = Distance from the source,

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$I_2 = I_1 \frac{r_1^2}{r_2^2}$

$I_2 = (0.10W/m^2)(\frac{(1m)^2}{(30m)^2})$

$I_2 = 1.11*10^{-4}W/m^2$

Now the expression for the Area of ear drum is

$A= \frac{\pi d^2}{4}$

$A = \frac{\pi (8.4mm\frac{1m}{10^3mm})^2}{4}$

$A = 5.5*10^{-5}m^2$

The expression for Sound power is

$P = AI_2$

Replacing,

$P= (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}W = 6.1*10^{-9}J/s$

Therefore the energy reached per second is $6.1*10^{-9}J$

4 0

3 years ago

A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185N at

Vitek1552 [10]

Answer:

$2.75 m/s^2$

Explanation:

We can solve the problem by writing the equations of motion along the horizontal and vertical direction.

Along the horizontal direction we have:

$T cos \theta - \mu N = ma$ (1)

where

$T cos \theta$ is the horizontal component of the tension, where

T = 185 N is the tension in the rope

$\theta=25^{\circ}$ is the angle between the rope and the horizontal

$\mu N$ is the force of friction, where

$\mu=0.27$ is the coefficient of friction

N is the normal reaction of the floor on the box

m = 35.0 kg is the mass of the box

a is the acceleration

Along the vertical direction we have:

$N+T sin \theta-mg=0$ (2)

where

N is the normal force (upward direction)

$T sin \theta$ is the vertical component of the tension in the rope (upward direction)

$mg$ is the weight of the box (downward direction), where

m = 35.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

From eq.(2) we get:

$N=mg-T sin \theta$

And substituting into (1), we can find the acceleration:

$T cos \theta - \mu (mg-T sin \theta) = ma\\Tcos \theta -\mu mg + \mu T sin \theta = ma\\a=\frac{T cos \theta- \mu mg + \mu T sin \theta}{m}=\\=\frac{(185)(cos 25^{\circ})-(0.27)(35.0)(9.8)+(0.27)(185)(sin 25^{\circ})}{35.0}=2.75 m/s^2$

8 0

3 years ago

How much work is done in accelerating a 2000 kg car from rest to a speed of 30 m/s?

Otrada [13]

The work done is the same as the amount of energy increase. The formula for kinetic energy is 122
1
2
m
v
2
.

The initial KE of the car is 12(1000)×202=200,000
1
2
(
1000
)
×
20
2
=
200
,
000
joules.

The final KE of the car is 12(1000)×302=450,000
1
2
(
1000
)
×
30
2
=
450
,
000
joules.

The difference between these is the amount of work done: 450,000−200,000=250,000
450
,
000
−
200
,
000
=
250
,
000
joules.
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8 0

3 years ago

HELP PLS

const2013 [10]

Answer:

Explanation:

The formula to be used here is that of the general gas law which is below

P₁V₁/T₁ = P₂V₂/T₂

where P₁ is the initial pressure (and can be represented as 1 atm)

V₁ is the initial volume (and can be represented as 1 cm³)

T₁ is the initial temperature (and can be represented as 1 °C = 274K)

P₂ is the final pressure (and would be 10 atm since the pressure increased 10 times)

T₂ is the final temperature (and would be 0.07°C = 273.07K since the temperature decreased 14 times; 1/14 = 0.07)

V₂ is the final volume which is unknown

Using the formula stated earlier

1 × 1/274 = 10 × V₂/273.07

1 × 1 × 273.07 = 10 × V₂ × 274

V₂ = 273.07/2740

V₂ = 0.099 ⇔ 0.1

Thus V₁:V₂ is 1:0.1

This means the volume of gas reduced 10 times (1/10) the original volume

8 0

3 years ago