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love history [14]

3 years ago

6

What is one way you are using nonrenewable resources?

1 answer:

lutik1710 [3]3 years ago

6 0

Putting gas in your car is one way you are using non-renewable resources.

You could Reduce, Recycle and Re-use items. You could manage what you do with you're water. Reducing and reusing products cuts down on manufacturing pollution, just as the use of recycled instead of virgin materials prevents pollution in industrial processes. You could add food coloring to your toilet tank. You could also do this: use sprinkler heads that sprinkle mist out instead of droplets of water. to avoid losing water to evaporation.

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What compounds are you made of?

Vanyuwa [196]

I am made of literally THOUSANDS of compounds ... too many to list here.
But the one compound that's most abundant, and actually comprises almost
80% of my entire beautiful body, is the compound DiHydrogen Oxide, with
the molecular formula H₂O . This compound is commonly known as "water".

6 0

2 years ago

Read 2 more answers

Can someone please answer this, ill give you brainliest Would be very appreciated.

marshall27 [118]

Answer:

cohesive properties

Explanation:

The property of cohesion allows liquid water to have <u>no tension on the surface</u>.

7 0

1 year ago

Read 2 more answers

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

3 years ago

At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

Passing of B occurs at 4108.31 height.

6 0

3 years ago

A 58.5 kg sprinter starts a race with an acceleration of 4.40 m/s2. What is the net external force on him?

mote1985 [20]

Answer:

External force on him = 257.40 N

Equation is; F = ma (where 'f' is force, 'm' is mass and 'a' is acceleration)

Explanation:

The mass of the sprinter is 58.5 kg

His acceleration is 4.40 m/s²

According to Newton's second law of motion; F = ma

External force on the sprinter = 58.5 kg × 4.40 m/s² = 257.40 N

4 0

2 years ago