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OLga [1]
3 years ago
6

What is the difference in graph shape when it is postion vs.time vs. velocity vs. time graph. Using a motion detector.

Physics
1 answer:
lions [1.4K]3 years ago
3 0
Heheheezbsbshhshshshh
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Pls help! True or False?
jasenka [17]

Answer:

True

Explanation:

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7 0
3 years ago
Read 2 more answers
How many cubic feet are in a 55-gallon drum?
san4es73 [151]
1 gallon = 231 cubic inches
1 cubic foot = 1728 cubic inches

                       (55 gal) x (231 in³/gal) x (1 ft³/1728 in³)

                   =    (55 x 231 / 1728)  ft³

                   =         7.352 cubic feet       (rounded)  
5 0
4 years ago
base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to laun
Nadya [2.5K]

Answer: v = 2.24 m/s

Explanation: The <u>Law</u> <u>of</u> <u>Conservation</u> <u>of</u> <u>Energy</u> states that total energy is constant in any process and, it cannot be created nor destroyed, only transformed.

So, in the toy launcher, the energy of the compressed spring, called <u>Elastic</u> <u>Potential</u> <u>Energy (PE)</u>, transforms into the movement of the plastic sphere, called <u>Kinetic</u> <u>Energy (KE)</u>. Since total energy must be constant:

KE_{i}+PE_{i}=KE_{f}+PE_{f}

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}

\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}

\frac{1}{2}50(0.1)^{2}=\frac{1}{2}(0.1)v^{2}

v^{2}=\frac{50(0.1)^{2}}{0.1}

v=\sqrt{50(0.1)}

v=\sqrt{5}

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

3 0
3 years ago
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0
3 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

4 0
3 years ago
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