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stiks02 [169]
1 year ago
12

If you speed through a construction zone while workers are present, your fines could be:.

Physics
1 answer:
valina [46]1 year ago
6 0

If you speed through a construction zone while workers are present, your fines could be as much as one thousand dollars.

<h3>What is a Fine?</h3>

This is referred to as the amount which is instructed by a court or authority to be paid as a result of it being a penalty for various types of offences. each crime has its specific fine which helps to serve as deterrent for unlawful behavior in the community.

it is always best not to speed when within a construction zone which has workers present in the area. This is ideal as it helps to prevent incidences of accidents or death.

it is therefore the reason why a fine of 1000 dollars is to be paid so that people can think of such high amount before performing certain types of activities when driving and makes it the most appropriate choice.

Read more about Fine here brainly.com/question/25229258

#SPJ1

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if an object is moving at a constant speed in a changing direction, What is the acceleration of the object (zero or not zero)​
Andreas93 [3]
  • We know, acceleration is the change of velocity by time.
  • Velocity is the speed of an object which also indicates the direction.
  • Hence, acceleration is both dependant upon the speed as well as the direction.
  • So, if an object is moving at a constant speed in a changing direction, the acceleration will also change. It will not be zero.
  • An example is that of uniform circular motion.

Answer:

if an object is moving at a constant speed in a changing direction, the acceleration of the object will not be zero.

4 0
2 years ago
A paper-filled capacitor is charged to a potential difference of V0=2.5 V. The dielectric constant of paper is k=3.7 . The capac
Pavlova-9 [17]

Answer:

k =  6.72

Explanation:

K of paper = 3.7

k of air = 1

Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So

V = Q / C = 2.5

Capacity becomes C / 3.7 in air .

capacity becomes C/3.7 when paper is replaced by air .

V₁ = Q / (C/3.7)

= 3.7 Q/C

3.7 x 2.5

= 9.25 V

In the second case ,

capacitance  due to new unknown dielectric k

= C/3.7 x k

= kC / 3.7 ( Capacitance in air is C/3.7 )

V ( new ) = Q / ( kC/3.7 )

= 3.7 Q/kC

.55 x 2.5 = 3.7 x( 2.5 / k )

k = 3.7 / .55

= 6.72

7 0
3 years ago
A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if
77julia77 [94]
C because when the part gets out of the probe it would no longer stay contacted
6 0
3 years ago
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
What would make oppositely charged objects attract each other more?
Lady_Fox [76]

Answer:

A. increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object

Explanation:

4 0
3 years ago
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