-- the big flash of light and heat coming out of the head
of a match when it gets hot enough;
-- the explosion of a tiny bit of gunpowder that can send
a bullet many miles;
-- the energy captured from a few drops of burning gasoline
that moves a car;
-- the energy in the carbohydrates you eat that is used
to move you around;
Answer:
r₁/r₂ = 1/2 = 0.5
Explanation:
The resistance of a wire is given by the following formula:
R = ρL/A
where,
R = Resistance of wire
ρ = resistivity of the material of wire
L = Length of wire
A = Cross-sectional area of wire = πr²
r = radius of wire
Therefore,
R = ρL/πr²
<u>FOR WIRE A</u>:
R₁ = ρ₁L₁/πr₁² -------- equation 1
<u>FOR WIRE B</u>:
R₂ = ρ₂L₂/πr₂² -------- equation 2
It is given that resistance of wire A is four times greater than the resistance of wire B.
R₁ = 4 R₂
using values from equation 1 and equation 2:
ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²
since, the material and length of both wires are same.
ρ₁ = ρ₂ = ρ
L₁ = L₂ = L
Therefore,
ρL/πr₁² = 4ρL/πr₂²
1/r₁² = 4/r₂²
r₁²/r₂² = 1/4
taking square root on both sides:
<u>r₁/r₂ = 1/2 = 0.5</u>
Answer:
Resistance of the second wire is twice the first wire.
Explanation:
Let us first see the formula of resistance;
R = pxL/A
Here L is the lenght of the wire, A the area and p is the resistivity of wire.
As we are given that the length of second wire is double than that of the first wire, hence the resistance of second wire would be double.
Since we have two loop in second case, inducing double voltage but as resistance is doubled so the current would remain same according to ohms law
I = V/R
Answer:
The law of refraction states that the incident ray, the refracted ray, and the normal to the interface, all lie in the same plane.
Explanation:
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J
