Repeat the experiment to make sure it gives the same results.
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r₁ = distance of point A from charge q₁ = 0.13 m
r₂ = distance of point A from charge q₂ = 0.24 m
r₃ = distance of point A from charge q₃ = 0.13 m
Electric field by charge q₁ at A is given as
E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C towards right
Electric field by charge q₂ at A is given as
E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C towards left
Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction
Electric field at A will be zero when
E₁ = E₂ + E₃
1.225 = 0.703 + E₃
E₃ = 0.522 N/C
Electric field by charge "q₃" is given as
E₃ = k q₃ /r₃²
0.522 = (9 x 10⁹) q₃/(0.13)²
q₃ = 0.980 x 10⁻¹² C = 0.980 pC
Answer:
.5 grams
Explanation:
1 gram is equal to 1000 milligrams (mg)
26.54 m/s is the magnitude of its velocity just before it strikes the ground
h=100m,v=20m/s,g=9.8m/s
time it takes to reach the ground,
![[t=\sqrt2h/g],[=\sqrt2*100/9.8=4.51s]](https://tex.z-dn.net/?f=%5Bt%3D%5Csqrt2h%2Fg%5D%2C%5B%3D%5Csqrt2%2A100%2F9.8%3D4.51s%5D)
x= 120m
t= 4.52
v= x/t
v= 120/4.52
v= 26.54 m/s
The "speed at which an object changes its location" can be expressed using a vector number called velocity. Consider a person who moves swiftly while taking two steps forward and two steps back while remaining in one location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A speed of 55 mph is not enough information. The direction must be used to appropriately depict the item's velocity. Simply said, the direction of the velocity vector indicates the direction of motion of an object.
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Answer:
W / A = 39200 kg / m²
Explanation:
For this problem let's use the equilibrium equation of / newton
F = W
Where F is the force of the door and W the weight of water
W = mg
We use the concept of density
ρ = m / V
m = ρ V
The volume of the water column is
V = A h
We replace
W = ρ A h g
On the other side the cylinder cover has a pressure
P = F / A
F = P A
We match the two equations
P A = ρ A h g
P = ρ g h
P = 39200 Pa
The weight of the water column is
W = 1000 9.8 4 A
W / A = 39200 kg / m²