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2 years ago

What is the relationship between environmental science and public policymakers?

1 answer:

4 0

In a simple sentence I would say using environmental science public policy maker would be able to create policies on environmental issues such as littering,recycling specific items/reusing,not wasting excess energy,ect.

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What value do we use to describe acceleration due to gravity

andrey2020 [161]

Answer:

9.8 m/s2

Explanation:

In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.

Got it from the internet, hope it helps though ^^

7 0

2 years ago

What is NOT a principle of genetics?

timurjin [86]

The recessive trait will always show up

3 0

2 years ago

Read 2 more answers

Which statement(s) is/are indicative of a covalent bond?

Over [174]

Do you have any answer options?

The most I can help is to tell you that a covalent bond is a type of bond that shares electrons between the bonded atoms.

4 0

3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

2 years ago

100 kw of power is delivered to the other side of a city by a pair of power lines with the voltage difference of 13014.1 v.

FinnZ [79.3K]

A) The power delivered to the lines is
$P_{in}= 100 kW=1 \cdot 10^5 W$
And the voltage at which the lines work is
$V=13014.1 V$
Since the power delivered is the product between the voltage and the current:
$P=VI$
We can find the current flowing in the lines:
$I= \frac{P}{V}= \frac{1 \cdot 10^5 W}{13014.1 V}=7.68 A$

b) The voltage change along each line can be found by using Ohm's law:
$\Delta V = IR = (7.68 A)(10 \Omega)=76.8 V$

c) The power wasted as heat along each line is given by:
$P_d = I^2 R = (7.68 A)^2 (10 \Omega) = 590 W$
And since we have 2 lines, the total power wasted as heat in both lines is
$P_d = 2 \cdot 590 W=1180 W$

6 0

2 years ago