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2 years ago

My uncle wants to play twister with me and my little sister but he gets a lil too wild. What do I do ?

Physics

2 answers:

Lelu [443]2 years ago

8 0

Get even wilder so he knows your'e not the one to joke with!

Hope This Helps Him Calm Down:)

NeTakaya2 years ago

7 0

Ask him not to be so wild. Hope this helps!

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Given the equation E = P/N , solve for P

Mars2501 [29]

Answer:

E=P/N

multiply both sides by N

P=EN

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2 years ago

(50 POINTS) please answer each question:

Advocard [28]

Answer:

i know the first one.

Explanation:

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2 years ago

Gasoline burns inside a car’s engine. how does this fuel enable a car to move?

Svetllana [295]

<span> The purpose of a gasoline car engine is to convert gasoline into motion so that your car can move. Currently the easiest way to create motion from gasoline is to burn the gasoline inside an engine.
Therefore, a car engine is an internal combustion engine -- combustion takes place internally.
There is such a thing as an external combustion engine. A steam engine in old-fashioned trains and steam boats is the best example of an external combustion engine. The fuel (coal, wood, oil, whatever) in a steam engine burns outside the engine to create steam, and the steam creates motion inside the engine. Internal combustion is a lot more efficient (takes less fuel per mile) than external combustion, plus an internal combustion engine is a lot smaller than an equivalent external combustion engine. This explains why we don't see any cars using steam engines.

To understand the basic idea behind how a reciprocating internal combustion engine works, it is helpful to have a good mental image of how "internal combustion" works.

One good example is an old Revolutionary War cannon. You have probably seen these in movies, where the soldiers load the cannon with gun powder and a cannon ball and light it. That is internal combustion, but it is hard to imagine that having anything to do with engines.

A potato cannon uses the basic principle behind any reciprocating internal combustion engine: If you put a tiny amount of high-energy fuel (like gasoline) in a small, enclosed space and ignite it, an incredible amount of energy is released in the form of expanding gas. You can use that energy to propel a potato 500 feet. In this case, the energy is translated into potato motion. You can also use it for more interesting purposes. For example, if you can create a cycle that allows you to set off explosions like this hundreds of times per minute, and if you can harness that energy in a useful way, what you have is the core of a car engine! </span>

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3 years ago

Different _______ of light through two separate mediums causes the bending of wave fronts associated with light rays.

eimsori [14]

The different reflections of light through two separate mediums causes the bending of wave fronts associated with light rays. The reflection and refraction is caused by the medium associated with its light rays.

3 0

3 years ago

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

3 years ago