Total distance = 76+54 = 130km
total time = 2+5 = 7hrs
Av. speed = 130/7 = 18.571km/hr = 18.6 km/h ( 3 sig fig)
Answer:
the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Explanation:
Given that;
final velocity v = 0
initial velocity u = 15m/s
time taken t = 4 s
acceleration a = ?
from the equation of motion
v = u + at
we substitute
0 = 15 + a × 4
acceleration a = -15/4 = - 3.75 m/s²
the negative sign tells us that its a deacceleration so the sign can be ignored.
Deacceleration due to friction a = μ × g
we substitute
3.75 = μ × 9.8
μ = 3.75 / 9.8 = 0.3826 ≈ 0.38
Therefore the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Answer:
T_final = 279.4 [°C]
Explanation:
In order to solve this problem, we must use the following equation of thermal energy.

where:
Q = heat = 9457 [cal]
m = mass = 79 [g] = 0.079 [kg]
Cp = specific heat = 0.5 [cal/g*°C]
T_initial = initial temperature = 40 [°C]
T_final = final temperature [°C]
![9457 = 79*0.5*(T_{f}-40)\\239.41=T_{f}-40\\\\T_{f}=279.4[C]](https://tex.z-dn.net/?f=9457%20%3D%2079%2A0.5%2A%28T_%7Bf%7D-40%29%5C%5C239.41%3DT_%7Bf%7D-40%5C%5C%5C%5CT_%7Bf%7D%3D279.4%5BC%5D)
To solve this problem we will apply the concept related to the electric field defined from the laws of Coulomb. For this purpose we will remember that the electric field is equivalent to the product of the Coulomb constant due to the change of the charge over the squared distance, mathematically this is

Here,
k = Coulomb's constant
r = Distance from center of terminal to point where electric field is to found
q = Excess charge placed on the center of terminal of Van de Graff's generator
Replacing we have that,


Therefore the electric field is 
<span>the dependent variable
<span>a variable (often denoted by y ) whose value depends on that of another.</span>
</span>