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Burka [1]
3 years ago
8

Your go-cart breaks down right before the end of a race, so you have to push it over the finish line. The go-cart has a mass of

85 kg.
a. What is the weight of your go-cart?

b. How much force must you apply to give the go-cart an acceleration of 0.9 m/s^2

c. If you push with a force of 350 N, what is the acceleration of the cart?

d. Later, you are driving your friend's go-cart. If his cart has an acceleration of 5.5 m/s^2 and the cart's engine is applying a forward force of 720 N, what is the mass of his go cart?
Physics
1 answer:
Flura [38]3 years ago
6 0

a) W=mg=833 N by definition

b) F=ma=76.5 N according to Newton's second law

c) a=F/m=4.12 m/s^2

d) m=F/a=720/5.5=130.91 kg

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What can you conclude about the total mechanical energy of a pendulum as it swings back and forth?
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Answer:

The total mechanical energy of a pendulum is conserved neglecting the friction.

Explanation:

  • When a simple pendulum swings back and forth, it has some energy associated with its motion.
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  • The potential energy of the simple pendulum is given by P.E = mgh
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What type of tool seems to be the mist accurate for measuring liquids?
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Volumetric flasks are most accurate

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Answer:

  1. 0.0121
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Explanation:

1.

\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{0.01223}{1.01}\:\\\mathrm{is}\:0.0121

2.

\frac{\begin{matrix}\space\space&\textbf{\space\space}&\space\space&\space\space&\space\space&4&9&10\\ \space\space&\textbf{1}&9&.&8&\linethrough{5}&\linethrough{10}&\linethrough{0}\\ -&\textbf{0}&0&.&0&1&1&3\end{matrix}}{\begin{matrix}\space\space&\textbf{1}&9&.&8&3&8&7\end{matrix}}\\\\=19.8387

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0.1886\times 12\\\\Multiply\:without\:the\:decimal\:points,\:then\:put\:the\:decimal\:point\:in\:the\:answer\\1886\times\:12=22632\\\\0.1886\mathrm{\:has\:}4\mathrm{\:decimal\:places}\\12\mathrm{\:has\:}0\mathrm{\:decimal\:places}\\\\\mathrm{Therefore,\:the\:answer\:has\:}4\mathrm{\:decimal\:places}\\\\=2.2632

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3 years ago
Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is
Elan Coil [88]

Answer:

The intensity of light from the 1mm from the central maximu is  I = 0.822I_o

Explanation:

From the question we are told that

                         The wavelength is \lambda = 620 nm = 620 *10^{-9}m

                         The width of the slit is w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m  

                          The distance from the screen is  D = 3.00m

                           The intensity at the central maximum is I_o

                          The distance from the central maximum is d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m

        Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

                     I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2

    Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

            sin \theta = \frac{z}{D}

           

if the angle between the the light at z and the central maximum is small

Then  sin \theta =  \theta

   Which implies that

              \theta = \frac{z}{D}

substituting this into the equation for the intensity

             I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D}  ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]

given that z =1mm = 1*10^{-3}m

   We have that

              I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2

                 =I_o [\frac{sin(0.760)}{0.760}] ^2

                 I = 0.822I_o

               

 

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3 years ago
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