Answer:
F = 7.68 10¹¹ N, θ = 45º
Explanation:
In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges
The net force is
F_ {net} = F₂₁ + F₂₃ + F₂₄
bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.
let's use trigonometry
cos 45 = F₂₄ₓ / F₂₄
sin 45 = F_{24y) / F₂₄
F₂₄ₓ = F₂₄ cos 45
F_{24y} = F₂₄ sin 45
let's do the sum on each axis
X axis
Fₓ = -F₂₁ + F₂₄ₓ
Fₓ = -F₂₁₁ + F₂₄ cos 45
Y axis
F_y = - F₂₃ + F_{24y}
F_y = -F₂₃ + F₂₄ sin 45
They indicate that the magnitude of all charges is the same, therefore
F₂₁ = F₂₃
Let's use Coulomb's law
F₂₁ = k q₁ q₂ / r₁₂²
the distance between the two charges is
r = a
F₂₁ = k q² / a²
we calculate F₂₄
F₂₄ = k q₂ q₄ / r₂₄²
the distance is
r² = a² + a²
r² = 2 a²
we substitute
F₂₄ = k q² / 2 a²
we substitute in the components of the forces
Fx = 
Fx = 
( -1 + ½ cos 45)
F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)
We calculate
F₀ = 9 10⁹ 4.25² / 0.440²
F₀ = 8.40 10¹¹ N
Fₓ = 8.40 10¹¹ (½ 0.707 - 1)
Fₓ = -5.43 10¹¹ N
remember cos 45 = sin 45
F_y = - 5.43 10¹¹ N
We can give the resultant force in two ways
a) F = Fₓ î + F_y ^j
F = -5.43 10¹¹ (i + j) N
b) In the form of module and angle.
For the module we use the Pythagorean theorem
F = 
F = 5.43 10¹¹ √2
F = 7.68 10¹¹ N
in angle is
θ = 45º
Answer:
i. 6.923 V
ii. The e.m.f. = 22.5 V
Explanation:
i. The given parameters are;
Length of potentiometer = 1 m
The resistance of the potentiometer = 10 Ω
The e. m. f. of the attached cell = 9 V
The current, I flowing in the circuit = e. m. f/(Total resistance)
The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A
The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire
The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V
ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;
Where:
E = e.m.f. of the balance point cell
= Resistance of 75 cm of potentiometer wire = 0.75×10 = 7.5 Ω
= Resistance of the cell in the circuit = 3 Ω
V = e.m.f. attached cell = 9 V
E = 7.5*3 = 22.5 V
The e.m.f. = 22.5 V
Answer:20°
Explanation:
Recall
Range R of a projectile is given by U^2sin2A/g
We're U = velocity,A= angle of projection and g is acceleration due to gravity
From the question the range R are the same
Hence R1=R2
U1^2sin2A/g=U2^2sin2B/g
But U1=U2 and g=g
Hence sin2A=sin 2B
Sin 2*70= sin2*B
0.6427=sin2B
B=sin inverse(0.6427)=40/2=20°
Answer:
Explanation:
We are given that
Mass of box=60kg
Net external force applied on the box=90 N
Friction force =30N
We have to find the acceleration of the box.
We know that
Net external force=ma
Substitute the values then we get
Hence, option b is true.
Both balls hit the ground at the same time.
We've known this fact for about 500 years (since Galileo),
and we've known WHY for about 300 years (since Newton).
