Answer:
Explanation:
Given
Force P is acting upward
C is vertical contact Force
W is the weight of the crate
As P is unable to move the Block therefore Normal reaction keeps on acting on block
thus we can say that
P-W+C=0
P=W-C
A soap box derby car at the top of a ramp has more _________ energy and at the finish line at the bottom of the ramp it has more
1 answer:
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Answer:
a) a = g / 3
b) x (3.0) = 14.7 m
c) m (3.0) = 29.4 g
Explanation:
Given:-
- The following differential equation for (x) the distance a rain drop has fallen has the form:
- Where, v = Speed of the raindrop
- Proposed solution to given ODE:
v = a*t
Where, a = acceleration of raindrop
Find:-
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.
Solution:-
- We know that acceleration (a) is the first derivative of velocity (v):
a = dv / dt ... Eq 1
- Similarly, we know that velocity (v) is the first derivative of displacement (x):
v = dx / dt , v = a*t ... proposed solution (Eq 2)
v .dt = dx = a*t . dt
- integrate both sides:
∫a*t . dt = ∫dt
x = 0.5*a*t^2 ... Eq 3
- Substitute Eq1 , 2 , 3 into the given ODE:
0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2
= 1.5 a^2 t^2
a = g / 3
- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:
x (t) = 0.5*a*t^2
x (t = 3.0) = 0.5*9.81*3^2 / 3
x (3.0) = 14.7 m
- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:
m (t) = k*x (t)
m (3.0) = 2.00*x(3.0)
m (3.0) = 2.00*14.7
m (3.0) = 29.4 g
Net flux through the cylindrical surface is given as
here q = enclosed charge in the surface
so here in order to find the value of q
so now we have
so this is the total flux
now by Gauss's law we can find the electric field
<em>by above expression we can find the electric field at required position</em>