Answer:
<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>
- <u><em>Initial velocity = 33 m/s</em></u>
- <u><em>Final velocity = 0 m/s</em></u>
- <u><em>Average velocity = (33 + 0) / 2 m/s = 16.5 m/s</em></u>
Explanation:
- <u><em>First, how long does it take the truck to come to a complete stop?</em></u>
- <u><em>( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds</em></u>
- <u><em>Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.</em></u>
Answer:
Net charge contained in the cubeq= 3.536×10^-6C
Explanation:
Formular for total flux in a cube is given as:
Total flux= E300Acos(180) + E200Acos(0)
Where A is crossectional area
Total flux= A(E200-E300)
Total flux= q/Eo
q= Eo×total flux
q=(8.84×10^-12)×(100)^2×(100-60)
q= 3.536×10^-6C
Answer:
The solution and the graph are in the attached files below
Explanation:
The weightiness of the added
water displaced is equivalent to the joined weight of the two extra people who come
to be into the boat:
<span>m water g = 2 x 690 N</span>
<span> =
1,380 N</span>
<span>
</span>
The mass of the water displace
is then
<span>m water g = 1,380 N</span>
<span> = 1,380 N / 9.8 m/s^2</span>
<span> = 141 kg</span>
<span>
</span>
Compute the calculation for
density for the volume of water displace and practice this outcome for the mass
of the water displace to get the answer:
<span>p water = mass of water / volume of water</span>
<span>
</span>
<span>volume of water = mass of water / p water</span>
<span> = 141 kg / 1000 kg /m^3 eliminate
kilogram</span>
<span> = 0.14 m^3 the additional volume
of water that is displaced</span>
