Answer:
0 - 60 mph = 0 - 26.8 m/s = 0 - 96.6 km/h; 0 - 100 km/h = 0 - 27.8 m/s = 0 - 62.1 mph.
Explanation:
Answer:
B. the number of field lines on the source charge
Explanation:
As we know that electric flux is defined as the number of electric field lines passing through a given area.
So here electric flux due to a point charge "q" is given by
so here we know that flux depends on the magnitude of charge and hence we can say that number of filed lines originating from a point charge will depends on the magnitude of the charge.
<span>You can use the equation
V_xf = V_xi + a_x(t)
V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0
t
Thus, solve for t and get 10seconds
and then take 5 seconds to break after 20 seconds of driving
so for
a) 10 + 20 + 5 = 35 seconds
</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
Need to find xf, knowing xi = 0
Thus, use the formula
x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
x_f = 100m
so for the first 10 seconds the truck traveled 100ms
At a speed of 20m/s
20m/s = xm/20s
20*20 = x
x = 400
thus we have 100+400 = 500m
then it slows down from 500m to x_f
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5)
x_f = 500 + 50
x_f = 550
therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s
thus
V_xavg = 16m/s</span>
Answer:
Explanation:
Calories to be burnt = 3500 - 2500 = 1000 Cals .
Efficiency of conversion to mechanical work is 25 % .
Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.
4000 Cals = 4.2 x 4000 = 16800 J .
Work done in one jump = kinetic energy while jumping
= 1/2 m v²
= .5 x 70 x 3.3²
= 381.15 J .
Number of jumps required = 16800 / 381.15
= 44 .
