Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.
Explanation: Hope this helped :)
Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
Answer:The term atomic number, conventionally denoted by the symbol Z, indicates number of protons present in the nucleus of an atom, which is also equal to the number of electrons in an uncharged atom. The number of neutrons is represented by the neutron number (N)
Explanation:
Answer: 330.88 J
Explanation:
Given
Linear velocity of the ball, v = 17.1 m/s
Distance from the joint, d = 0.47 m
Moment of inertia, I = 0.5 kgm²
The rotational kinetic energy, KE(rot) of an object is given by
KE(rot) = 1/2Iw²
Also, the angular velocity is given
w = v/r
Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy
w = v/r
w = 17.1 / 0.47
w = 36.38 rad/s
Now, substituting the value of w, with the already given value of I in the equation, we have
KE(rot) = 1/2Iw²
KE(rot) = 1/2 * 0.5 * 36.38²
KE(rot) = 0.25 * 1323.5
KE(rot) = 330.88 J
Explanation:
the answer is 2.46 × 10^12
