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sesenic [268]
3 years ago
6

PLS HELP I NEED ANSWER HOMEWORK DUE SOON

Physics
1 answer:
vovangra [49]3 years ago
4 0

Answer:

I think its 1

Explanation:

The reason why I think it's 1 is because 0.5m/s^2 /0.5 m/s= 1 s

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Rearrange the formula for mechanical energy to solve for height:
docker41 [41]

Explanation:

Given formula:

            ME=\frac{1}{2}mv²+mgh

  To make height the subject of the formula, follow the following procedures;

     Subtract \frac{1}{2}mv² from both side of equation

 M.E - \frac{1}{2}mv² = \frac{1}{2}mv² - \frac{1}{2}mv²+mgh

                  This gives:

                        M.E - \frac{1}{2}mv² = mgh

Multiply both sides of the expression by \frac{1}{mg}

  ( M.E -  \frac{1}{2}mv² ) x  \frac{1}{mg} =  \frac{1}{mg} x mgh

       h = ( M.E -  \frac{1}{2}mv² ) x  \frac{1}{mg}

Learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

4 0
3 years ago
In a coiled spring, the particles of the medium vibrate to and fro about their
Kitty [74]

Answer:

In a coiled spring, the particles of the medium vibrate to and fro about their mean positions at an angle of

A. 0° to the direction of propagation of wave

Explanation:

The waveform of a coiled spring is a longitudinal wave, which is made up of vibrations of the spring which are in the same direction as the direction of the wave's advancement

As the coiled spring experiences a compression force and is then released, it experiences a sequential movement of the wave of the compression that extends the length of the coiled spring which is then followed by a stretched section of the coiled spring in a repeatedly such that the direction of vibration of particles of the coiled is parallel to direction of motion of the wave

From which we have that the angle between the direction of vibration of the particles of the coiled spring and the direction of propagation of the wave is 0°.

8 0
2 years ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
3 years ago
Can someone please help me with these. I don't understand any of it.
ValentinkaMS [17]
1. electrical energy: electrical energy that is caused by moving electrons

2. coolant: a mixture of antifreeze and water that removes excess heat from an internal engine

3. electric compressor: a device that <span>acts as a pump, circulating refrigerant throughout the refrigerator 

4. the inside of the fridge and the food becomes colder

5. the coolants becomes a hot, high-pressure gas 

6. as coolant transfers thermal energy to the air outside, it turns back into a liquid

</span>
6 0
3 years ago
How is electrolysis used in order to prevent materials from corrosion or rusting?​
KengaRu [80]

Answer:

We use electrolysis to prevent a material from rusting,

The metal forms a coating around the material and hence prevents any contact between the material and the environment

This process also gives us the physical strength of the material and the aesthetic properties of the coated metal

the metal commonly used to coat the object is Zinc and the process is called galvanisation

6 0
3 years ago
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