Applying the Newton Second Law of motion that is F=ma
We have F=395.2N (North) and m=259.1Kg
Putting these values in the equation
395.2 = 259.1 x a than
for a = 395.2/259.1
a = 1.525 m/sec 2
Radium........ Hope this helps :)
Answer:
Temperature of the hot reservoir is 1540K
Explanation:
![E= 1- \frac{T_{c}}[tex]{T_h}=308+{T_c}\\Efficiency of a carnot engine is given by the aboveTc=temperature of the cold reservoirTh= temperature of the hot reservoirK=273+ 35 (convert 35°C to kelvin)K=308k{T_h}={T_c}+308-----------------------(equation 1)20%=1-{T_c}/{T_h}](https://tex.z-dn.net/?f=E%3D%201-%20%5Cfrac%7BT_%7Bc%7D%7D%5Btex%5D%7BT_h%7D%3D308%2B%7BT_c%7D%5C%5C%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EEfficiency%20of%20a%20carnot%20engine%20is%20given%20by%20the%20above%3C%2Fp%3E%3Cp%3ETc%3Dtemperature%20of%20the%20cold%20reservoir%3C%2Fp%3E%3Cp%3ETh%3D%20temperature%20of%20the%20hot%20reservoir%3C%2Fp%3E%3Cp%3EK%3D273%2B%2035%20%20%28convert%20%2035%C2%B0C%20to%20kelvin%29%3C%2Fp%3E%3Cp%3EK%3D308k%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%7BT_h%7D%3D%7BT_c%7D%2B308-----------------------%28equation%20%201%29%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3E20%25%3D1-%7BT_c%7D%2F%7BT_h%7D)
0.2=({T_c}+308-{T_c})/{T_c}+308
.2({T_c}+61.6=308
0.2{T_c}=246.4
{T_c}=1232
recall from equation 1
{T_h}=308+1232
{T_h}=1540K
Answer:
acceleration of the car is 3 m\s^2
Explanation:
from rest means the initial velocity (vi) is zero
time = 5s
final velocity (vf) = 15m\s
a = vf - vi \ t
a = (15-0) \ 5
a= 3 m\s^2
which means that the car is speeding up 3 meters every second
<span>the speed of something in a given direction. so i think none of these</span>
