Answer:
Explanation:
Let c be the circumference and r be the radius
c = 2πr , r = c / 2π , area A = π r² = π (c/2π )² = (1/4π) x c²
flux (ψ) = BA = 1 X 1/4π X c²
dψ/dt = 1/4π x 2c dc/dt =1/2π x c x dc/dt
at t = 8 s
c = 161 - 13 x 8 = 57 cm , dc/dt = 13 cm/s
e = dψ/dt = (1 / 2π )x 57 x 13 x 10⁻⁴ = 118 x 10⁻⁴ V.
Explanation:
The mass written on the periodic table is an average atomic mass taken from all known isotopes of an element. This average is a weighted average, meaning the isotope's relative abundance changes its impact on the final average. The reason this is done is because there is no set mass for an element.
Answer:
2.87m
Explanation:
Using the law of gravitation to solve this question
F = GMm/r²
G is the gravitational constant
M and m are the masses
r is the distance between the masses
Substitute the given values
G = 6.67×10^-11 m³/kgs²
M =8.8 x 10^6 kg
m = 5.6 x 10^5 kg
F =440N
400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²
400r² = 328.698×10
400r² = 3286.98
r² = 3286.98/400
r² = 8.21745
r = √8.21745
r = 2.87m
Hence the distance of separation is 2.87m
Answer:
W = - 5.01 10¹⁰ J
Explanation:
Work is defined by the expression
W = ∫ F.dr
Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product
W = ∫ F dr
W = G m₁m₂ ∫ 1 /r² dr
W = G m₁ m₂2(-1 / r)
We evaluate between the lower limits r = Re and upper r = ∞
W = G m₁m₂ (-1 / Re + 1 / ∞)
W = - G m₁ m₂ / Re
Let's calculate
W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶
W = - 5.01 10¹⁰ J
Answer:
The function that describe the motion in the time
y (t) = 0.28m * sin ( 36.025 * t)
Explanation:
The angular frequency of oscillation of the spring
w = √k/m
w = √305 N/m / 0.235 kg
w = 36.025 rad / s
To determine the function of the motion knowing as a motion oscillation in a amplitude a frequency
y(t) = A * sin (w t )
So
A = 28.0 cm * 1 m / 100 cm = 0.28 m
So replacing to determine the function of the motion in the time
y (t) = A sin (w t)
y (t) = 0.28m * sin ( 36.025 * t)
