Answer:
a=8.06m/s^2
Explanation:
The box can be considered negligible body slidding down along a curved path defined by the parabola Y=Ax^2
Note:
When it's at A(x=2m, y=1.6m),
the speed Vb=8m/s and the increase in speed=4m/s^2
To find the acceleration,
Y=Ax^2
dy/dx=8x
d^2y/dx^2=8
p={[1+(dy/dx)^2]^3/2}/|d^2y/dx^2| .......1
substituting into 1, we have
p=8.39624m
an=v^2/p
an=8^2/8.39624=7.6224m/s^2
a=sqrt(at^2+an^2)
a=sqrt(4^2+7.62246^2)
a=8.06m/s^2
Answer:
Explanation:
Left block is on surface with higher inclination so it will go down . If T be tension
For motion of block A ,
net force = mgsin60 - (T + mg cos 60 x μ ) , μ is coefficient of friction .
ma = mgsin60 - T - mg cos 60 x .1
10a = 277.13 - T - 16
= 261.13 - T
T = 261.13 - 10a
For motion of block B
T - mg sin30 - mgcos30 x μ = ma
T- 160 - 27.71 = 10 a
261.13 - 10a - 160 - 27.71 = 10a
73.42 = 20a
a = 3.67 ft / s²
common acceleration = 3.67 ft / s²
Answer:
Area = 20 m²
Explanation:
Given the following data;
Force = 2500 N
Pressure = 125 Pa
To find the area on which it rest;
Mathematically, pressure is given by the formula;
Making area the subject of formula, we have;
Substituting into the formula, we have;
Area = 20 m²
Answer:
0.012 J
Explanation:
We are given:
q = 0.0080C
Potential difference = 1.5V
W=qV
Substituting the values into the equation:
W=0.0080*1.5= 0.012J
Answer:
E = - dV / dx
Explanation:
The equipotential lines are lines or surfaces that have the same power, therefore we can move in them without carrying out work between equipotential lines, work must be carried out, therefore the electric field changes.
The electric field and the potential are related by
E = - dV / dx
therefore when the change is faster, that is, the equipotential lines are closer, the greater the electric field must be.