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crimeas [40]
3 years ago
12

Pins accompained by a ball bowling dates back to what era

Physics
2 answers:
Afina-wow [57]3 years ago
6 0
That is one of the weirdest questions
jekas [21]3 years ago
5 0

Answer:

Ancient Egypt

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Neo and Morpheus's masses have gained a velocity (not equal to zero) which means their momentum is now _____ .
Arte-miy333 [17]
<span>Neo and Morpheus's masses have gained a velocity (not equal to zero) which means their momentum is now based on gravity and friction alone.</span>
6 0
3 years ago
Read 2 more answers
What is the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25
alexandr1967 [171]

The mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

HOW TO CALCULATE MASS:

  • The mass of an object can be calculated by dividing the force applied to the object by its acceleration.

  • According to this question, a bus can accelerate from rest to 15.5 m/s over 8.25s. The acceleration can be calculated as follows:

  • a = (v - u)/t

  • a = 15.5 - 0/8.25

  • a = 15.5/8.25

  • a = 1.88m/s²

  • The mass of the bus = 7500N ÷ 1.88m/s²

  • The mass of the bus = 3989.4kg

  • Therefore, the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

Learn more about mass at: brainly.com/question/20259048?referrer=searchResults

3 0
3 years ago
A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the
MArishka [77]

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

Explanation:

From the question we are told that

   The  width of the slit is  D  =  0.3 \ mm =  0.3 *10^{-3} \ m

    The  wavelength is  \lambda =  254 \ nm =  254 *10^{-9} \ m

     The angle is  \theta  =  11^o

The intensity of at 11^o to the axis in terms of the intensity of the central maximum. is mathematically represented as

        I_c = \frac{I}{I_o}  = [ \frac{sin \beta  }{\beta }] ^2

Where \beta is mathematically represented as

        \beta  =  \frac{D sin (\theta ) *  \pi}{\lambda }

substituting values

      \beta  =  \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }

     \beta  =  708.1 \ rad

So

  I_c = \frac{I}{I_o}  = [ \frac{sin (708.1)  }{(708.1)}] ^2

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

6 0
4 years ago
References
never [62]

a b c d

A V=V+gt

dhehehdhehdhshehehehehehe

4 0
3 years ago
Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1
Tju [1.3M]

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

5 0
3 years ago
Read 2 more answers
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