Answer:
F = 69.3 N
Explanation:
For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by
fr = μ N
We define a reference system parallel to the floor
block B ( lower)
Y axis
N - W₁-W₂ = 0
N = W₂ + W₂
N = (M + m) g
X axis
F -fr = M a
for block A (upper)
X axis
fr = m a (2)
so that the blocks do not slide, the acceleration in both must be the same.
Let's solve the system by adding the two equations
F = (M + m) a (3)
a =
the friction force has the formula
fr = μ N
fr = μ (M + m) g
let's calculate
fr = 0.34 (2.0 + 0.250) 9.8
fr = 7.7 N
we substitute in equation 2
fr = m a
a = fr / m
a = 7.7 / 0.250
a = 30.8 m / s²
we substitute in equation 3
F = (2.0 + 0.250) 30.8
F = 69.3 N
Answer:
A. 5.4 * 10^(-4) m
B. 500V
Explanation:
A. Electric potential, V is given as:
V = kq/r
This means that radius, r is
r = kq/V
r = (9 * 10^9 * 30 * 10^(-12))/500
r = (270 * 10^(-3))/500
r = 5.4 * 10^(-4) m
B. Now the radius is doubled and the charge is doubled,
V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)
V = 500V
Answer:
Gravity
Explanation:
The answer is gravity because when the 3 masses were hung from the spring, gravity pulled the spring towards the ground.