Answer:
The tension in the string is quadrupled i.e. increased by a factor of 4.
Explanation:
The tension in the string is the centripetal force. This force is given by

m is the mass, v is the velocity and r is the radius.
It follows that
, provided m and r are constant.
When v is doubled, the new force,
, is

Hence, the tension in the string is quadrupled.
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
the first one is medium, the second one is type, and the third one is temperature
. if i gave the correct answer, please give best answer x
Answer
given,
net charge = +2.00 μC
we know,
1 coulomb charge = 6.28 x 10¹⁸electrons
1 micro coulomb charge = 6.28 x 10¹⁸ x 10⁻⁶ electron
= 6.28 x 10¹² electrons
2.00 μC = 2 x 6.28 x 10¹² electrons
= 1.256 x 10¹³ electrons
since net charge is positive.
The number of protons should be 1.256 x 10¹³ more than electrons.
hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.
The hoop is attached.
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is
61°.</span>