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4 years ago

Which phrases describe the atmosphere? Check all that apply.

Physics

1 answer:

vladimir2022 [97]4 years ago

7 0

Answer:

There is no options for this answer

Explanation:

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________ law is called the law of partial pressure

Studentka2010 [4]

Dalton's <span>law of partial pressures
</span>

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3 years ago

On a hike, you travel 2800 m in 2 h. What is your average speed?

Marina86 [1]

<h2>Greetings!</h2>

Firstly, to find speed you need to remember the speed formula:

Speed = Distance ÷ Time

You can do it by doing 2800 ÷ 2 but that gives m/h (metres per hour)

So to convert hours into seconds, you multiply 2 by 7200 (amount of second in 2 hours:

2 * 7200 = 14400 seconds

Now we can do distance ÷ time by plugging the values in:

2800 ÷ 14400 = $\frac{7}{36}$ ≈0.194

<h3>So the average speed is 0.194m/s (metres per second) </h3><h3 /><h2>Hope this helps!</h2>

7 0

3 years ago

Explain the energy transformations that occur when breaking in a hybrid<br> vehicle.

KatRina [158]

it occur kinetic energy

when breaking in a hybrid

vehicle.

hope it helps

8 0

3 years ago

Read 2 more answers

What does not affect the gravitational potential energy of an object

natta225 [31]

Anything really like velocity? as long as its not mass, gravity, and height

Ep = mgh

Those are the only variables that affect potential energy

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3 years ago

Read 2 more answers

A plastic block of mass 0.20.2kg is initially at rest on a horizontal frictionless surface. A bullet of mass 1212 g is fired wit

Juli2301 [7.4K]

Answer:

94.29619%

Explanation:

$m_1$ = Mass of bullet= 0.012 kg

$m_2$ = Mass of block = 0.2 kg

$u_1$ = Initial Velocity of bullet= 550 m/s

$u_2$ = Initial Velocity of block = 0 m/s

$v_1$ = Final Velocity of bullet = 20 m/s

$v_2$ = Final Velocity of block

As the linear momentum of the system is conserved

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{0.012\times 550+0.2\times 0-0.012\times 20}{0.2}\\\Rightarrow v_2=31.8\ m/s$

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(m_1u_1^2-m_1v_1^2-m_2v_2^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(0.012\times 550^2-0.012\times 20^2-0.2\times 31.8^2)\\\Rightarrow \Delta K=1711.476\ J$

Percentage change is given by

$\dfrac{\Delta K}{\dfrac{1}{2}m_1u_1^2}\times 100=\dfrac{1711.476}{\dfrac{1}{2}0.012\times 550^2}\times 100=94.29619\ \%$

The percent of the initial kinetic energy of the bullet block system is lost during the collision is 94.29619%

4 0

3 years ago