Answer:
-4*10⁴ units.
Explanation:
As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must still remain to be zero.
So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.
Answer:
I would use the model of Ammonia because it helps you visualize the structure of NH3 better than the description. It would be easier to understand the structure of it if you can see it, rather than reading its description.
Answer:
The period of motion of new mass T = 0.637 sec
Explanation:
Given data
Mass of object (m) = 9 gm = 0.009 kg
Δx = 3.5 cm = 0.035 m
We know that spring force is given by
F = m g
F = 0.009 × 9.81 = 0.08829 N
Spring constant
![k = \frac{F}{x}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7BF%7D%7Bx%7D)
![k = \frac{0.08829}{0.035}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B0.08829%7D%7B0.035%7D)
k = 2.522 ![\frac{N}{m}](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7Bm%7D)
New mass
= 26 gm = 0.026 kg
Now the period of motion is given by
![T = 2 \pi \sqrt{\frac{m}{k} }](https://tex.z-dn.net/?f=T%20%3D%202%20%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7D)
![T = 2 \pi \sqrt{\frac{0.026}{2.522} }](https://tex.z-dn.net/?f=T%20%3D%202%20%5Cpi%20%5Csqrt%7B%5Cfrac%7B0.026%7D%7B2.522%7D%20%7D)
T = 0.637 sec
This is the period of motion of new mass.
Answer:
499.523. meter
<em>I</em><em> hope</em><em> it's</em><em> helps</em><em> you</em>
Answer:
a) E = -4 10² N / C
, b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰
Explanation:
For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball
X axis
-
= m a
Axis y
- W = 0
Initially the system is in equilibrium, so zero acceleration
Fe =
T_{y} = W
Let us search with trigonometry the components of the tendency
cos θ = T_{y} / T
sin θ =
/ T
T_{y} = cos θ
= T sin θ
We replace
q E = T sin θ
mg = T cosθ
a) the electric force is
= q E
E =
/ q
E = -0.032 / 80 10⁻⁶
E = -4 10² N / C
b) the distance to this point can be found by dividing the two equations
q E / mg = tan θ
θ = tan⁻¹ qE / mg
Let's calculate
θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)
θ = tan⁻¹ 0.3265
θ = 18
⁰
sin 18 = x/0.30
x =0.30 sin 18
x = 0.093 m
c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations
X axis
= m aₓ
aₓ = q E / m
aₓ = 80 10⁻⁶ 4 10² / 0.01
aₓ = 3.2 m / s²
Axis y
W = m
a_{y} = g
a_{y} = 9.8 m/s²
The total acceleration is can be found using Pythagoras' theorem
a = √ aₓ² + a_{y}²
a = √ 3.2² + 9.8²
a = 10.31 m / s²
The Angle meet him with trigonometry
tan θ = a_{y} / aₓ
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-9.8) / 3.2
θ = -71.9⁰
Movement is two-dimensional type with acceleration in both axes