Answer:
a. ac = 1844.66 m/s²
b. Fc = 265.63 N
Explanation:
a.
The centripetal acceleration of the ball is given as follows:
ac = v²/r
where,
ac = centripetal acceleration = ?
v = speed of ball = (87 mph)(1 h/ 3600 s)(1609.34 m / 1 mile) = 38.9 m/s
r = radius of path = 82 cm = 0.82 m
Therefore,
ac = (38.9 m/s)²/0.82 m
<u>ac = 1844.66 m/s²</u>
<u></u>
b.
The centripetal force is given as:
Fc = (m)(ac)
Fc = (0.144 kg)(1844.66 m/s²)
<u>Fc = 265.63 N</u>
The answer is the third one down. New evidence may contradict the old evidence of a certain theory.
Hey there, Your answer wold be 1.36 newtons rounded to 1.4 newton. This is how!
force = k (q1 * q2) / r^2
<span>= (9.0 × 10^9 Nm^2/C^2) [(7.1 × 10^-6 C) (6.9 × 10^-6 C)] / (5.7 × 10^-1 m)^2
</span>
<span>= [ 9.0 × 7.1 × 6.9 / (5.7)^2 ] × 10^-1 N
</span>
<span>= 13.6 × 10^-1 N
</span>
<span>= 1.36 N
</span>
ANSWER-1.4
HOPE I HELPED!!!!!!!!!!
a. The force applied would be equal to the frictional
force.
F = us Fn
where, F = applied force = 35 N, us = coeff of static
friction, Fn = normal force = weight
35 N = us * (6 kg * 9.81 m/s^2)
us = 0.595
b. The force applied would now be the sum of the
frictional force and force due to acceleration
F = uk Fn + m a
where, uk = coeff of kinetic friction
35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)
uk = 0.533
Answer:
a) 
b)
Explanation:
Given that:
- mass of rod,
- length of the rod,
<u>(a)</u>
<u>Moment of inertia of rod about its center and perpendicular to the rod is given as:</u>
(b)
<u>Moment of inertia on bending the rod to V-shape of 60 degree angle and axis being perpendicular to the plane of V at the vertex.</u>
<em>We treat it as two rod with axis of rotation at the end and perpendicular to the plane of rotation. </em>
<em>So, the mass and the length of the rod will become half of initial value.</em>
![I=2[ \frac{1}{3}\times 0.2\times 0.3^2]](https://tex.z-dn.net/?f=I%3D2%5B%20%5Cfrac%7B1%7D%7B3%7D%5Ctimes%200.2%5Ctimes%200.3%5E2%5D)
