Answer:
<em>4.61 N</em>
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Explanation:
masa = 0.5 kg
ángulo de inclinación = 20°
Peso normal de la masa = mg
donde m = masa
g = aceleración debido a la gravedad = 9.81 m/s^2
Peso normal = 0.5 x 9.81 = 4.905 N
Si la masa se mantiene en su lugar mediante una cuerda paralela al plano, y no hay fricción en la masa, entonces
La fuerza sobre la cuerda = peso normal x cos ∅
donde ∅ = 20 °
La fuerza sobre la cuerda = 4.905 x cos 20°
==><em> 4.61 N</em>
Answer:
K = 2 10⁻⁸ J
Explanation:
Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor
C = Q / ΔV
C = ε₀ A / d
ε₀ A / d = Q / ΔV
Q = ε₀ A ΔV / d (1)
indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,
as the power supply is disconnected and the capacitor is ideal the charge remains constant
in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1
ΔV₂ =
we substitute the equation for Q
ΔV₂ =
ΔV₂ =
in the third part we use the concepts of energy
starting point. Test charge near positive plate
Em₀ = U = q ΔV₂
final point. Test charge near negative plate
Em_f = K
energy is conserved
Em₀ = Em_f
q ΔV₂ = K
K = q ΔV₁
we calculate
K = 1 10⁻⁹ 12 0.005/0.003
K = 2 10⁻⁸ J
We have to remember a point , which is ' the cart or spring rest on a smooth horizontal track ' , i.e., any frictional force doesn't take place.
<u>Explanation:</u>
a) According to the question the cart is pulled to position A and released, i.e., the velocity of the cart at A initially (say time,t=0) is 0 m/s ,then moves toward position E, where it reverses direction and returns again to position A , in the 2nd phase cart moves along A to E , the cart's velocity increase and again goes to zero at point E and again change the direction, hence
( File has been attached)
b) Let's , the distance between two consecutive points is x meter and the spring constant is k N.m
c) ( File has been attached)
d) Movinf Right
e) Moving left
Answer:
D) F
Explanation:
Let m and M be the mass of the balls A and B respectively and r be the distance between the two balls. The magnitude of attractive gravitational force experienced by the balls due to each other is given by the relation :
......(1)
Now, if the masses of both the balls gets doubled as well as there separation distance also gets doubled, then let F₁ be the new gravitational force acting on them.
Since, New mass of ball A = 2M
New mass of ball b = 2m
Distance between the two balls = 2r
Substitute these values in equation (1).
Using equation (1) in the above equation.
F₁ = F