Why does rubbing a balloons on wool or hair make it attract other objects
2 answers:
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Answer:
The potential difference between points A and B is 278.95 volts.
The potential difference between points B and C is -642.10 volts.
The potential difference between points A and C is -363.15 volts.
Explanation:
Given :
Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C
Work is done to move a charge from point A to B, W₁ = 5.3 μJ
Work done to move a charge from point B to C, W₂ = -12.2 μJ
Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.
The relation between work done and potential difference is:
W = qV
V = W/q ....(1)
Using equation (1), the potential difference between points A and B is:
Substitute the suitable values in the above equation.
V₁ = 278.95 V
Using equation (1), the potential difference between points B and C is:
Substitute the suitable values in the above equation.
V₂ = -642.10 V
The potential difference between points A and C is:
V₃ = V₁ + V₂
V₃ = 278.95 - 642.10
V₃ = -363.15 V
Answer:
B) 27.3 m
Explanation:
The rock describes a parabolic path.
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = vx*t Equation (1)
Where:
x: horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m/s
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:
(vfy)² = (v₀y)² - 2g(y- y₀) Equation (2)
vfy = v₀y -gt Equation (3)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Data
v₀ = 30 m/s , at an angle α=40.0° above the horizontal
v₀x = vx = 30*cos40° = 22.98 m/s
v₀y = 30*sin40° = 19.28 m/s
y₀ = 2m
y = 18.0 m
g = 9.8 m/s²
Calculation of the time (t) it takes for the rock to reach at 18 m above the ground
We replace data in the equation (2)
(vfy)² = (v₀y)² - 2g(y- y₀)
(vfy)² = (19.28)² - 2(9.8)(18- 2)
(vfy)² = 371.86 - 313.6
(vfy)² = 58.26
vfy = 7.63 m/s
We replace vfy = 7.63 m/s in the equation (2)
vfy = v₀y - gt
7.63 = 19.28 - (9.8)(t)
(9.8)(t) = 11.65
t = 11.65 / (9.8)
t = 1.19 s
Horizontal distance from where the rock was thrown to the window
We replace t = 1.19 s , in the equation (1)
x = vx*t
x = (22.98)* ( 1.19 )
x = 27.3 m