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4 years ago

Which of the following is an important step in the process of conflict resolution?

Physics

2 answers:

4vir4ik [10]4 years ago

7 0

A would be the correct answer. Its the only one to make sense since you are trying to solve the conflict!

Darya [45]4 years ago

6 0

The answer could be A

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Part C

iVinArrow [24]

Answer:the wave lengths are shorter and faster

Explanation:

Plato

7 0

3 years ago

A student drops a ball off the top of building and records that the ball takes 3.82s to reach the ground. Determine all unknowns

klemol [59]

Explanation:

Given parameters:

Time of drop = 3.82s

Unknown:

Final velocity of the ball = ?

Height of fall = ?

Solution:

To solve this problem, we apply the appropriate motion equation.

To find the final velocity;

v = u + gt

v is the unknown final velocity

u is the initial velocity = 0m/s

g is the acceleration due to gravity = 9.8m/s²

t is the time

v = 0 + 9.8 x 3.82 = 37.4m/s

Height of drop;

H = ut + $\frac{1}{2}$ gt²

So;

H = (0 x 3.82) + ( $\frac{1}{2}$ x 9.8 x 3.82²) = 71.5m

4 0

3 years ago

In the food web shown above, which of the following organisms can be classified as a tertiary consumer?

Anvisha [2.4K]

Hello,

Here is your answer:

The proper answer to this question is option A "puffin".

Your answer is A.

If you need anymore help feel free to ask me!

Hope this helps!

5 0

3 years ago

Read 2 more answers

A huge cannon is assembled on an airless planet (ignore any effects due to the planet's rotation). The planet has a radius of 5.

Ganezh [65]

Answer:

The projectile's speed as it passes the satellite is 1497.8 m/s.

Explanation:

Given that,

Radius of planet $r=5.00\times10^{6}\ m$

Mass of planet $m=3.95\times10^{23}\ kg$

Speed = 2000 m/s

Height = 1000 km

We need to calculate the projectile's speed as it passes the satellite

Using conservation of energy

$E_{1}=E_{2}$

$\dfrac{1}{2}mv_{1}^2+\dfrac{GmM}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{GmM}{R+h}$

$\dfrac{v_{1}^2}{2}+\dfrac{GM}{r_{1}}=\dfrac{v_{2}^2}{2}+\dfrac{GM}{R+h}$

$-\dfrac{v_{2}^2}{2}=-(\dfrac{GM}{R}-\dfrac{GM}{R+h}-\dfrac{v_{1}^2}{2})$

$v_{2}^2=v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})$

$v_{2}=\sqrt{v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})}$

Put the value into the formula

$v_{2}=\sqrt{2000^2+2\times6.67\times10^{-11}\times3.95\times10^{23}(\dfrac{1}{5.00\times10^{6}+1000\times10^{3}}-\dfrac{1}{5.00\times10^{6}})}$

$v_{2}=1497.8\ m/s$

Hence, The projectile's speed as it passes the satellite is 1497.8 m/s.

4 0

3 years ago

He went back to the video to see what had been recorded and was shocked at what he saw.

jeka94

So what’s the question?

5 0

3 years ago