Un objeto con una carga de 7μC se coloca en un campo eléctrico uniforme de 200N/C con dirección vertical. ¿Cuál es el valor de l
1 answer:
You might be interested in
The radius of the sphere in meters is ,r =
Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,
r/10 = tan [arctan(1/2)] = tan (1/2)
Since, tan (θ/2) = 1-cos(θ) / sin(θ)
we find that,
r/10 =
Hence, r =
So, the radius of the sphere in meters is ,r =
Learn more about radius (r) of the sphere here;
#SPJ4
This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
