Question is not complete and the missing part is;
A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.
Answer:
0.828 m/s
Explanation:
Resolving vertically, we have;
Fn and Fg act vertically. Thus,
Fn - Fg = 0 - - - - eq(1)
Resolving horizontally, we have;
Ff = ma - - - - eq(2)
Now, Fn and Fg are both mg and both will cancel out in eq 1.
Leaving us with eq 2.
So, Ff = ma
Now, Frictional force: Ff = μmg where μ is coefficient of friction.
Also, a = v²/r
Where v is linear speed or velocity
Thus,
μmg = mv²/r
m will cancel out,
Thus, μg = v²/r
Making v the subject;
rμg = v²
v = √rμg
Plugging in the relevant values,
v = √0.14 x 0.5 x 9.8
v = √0.686
v = 0.828 m/s
Answer:
1.07 x 10⁻⁸N
Explanation:
Given parameters:
Mass 1 = 200kg
Mass 2 = 500kg
Distance of separation = 25m
Unknown:
Gravitational attraction between the two bodies = ?
Solution:
To solve this problem, we use the equation of the universal gravitation;
F =
G is the universal gravitation constant = 6.67 x 10⁻¹¹Nm²kg⁻²
r is the distance
Now insert the parameters and solve;
F =
= 1.07 x 10⁻⁸N
Answer:
<em>The collision last 0.83 s</em>
Explanation:
Using the equation of motion,
S = (v+u)t/2 ........................ Equation 1
making t the subject of the equation,
t = 2S/(v+u)..................... Equation 2
Where S = distance, v = final velocity, u = initial velocity, t = time
Given: v = 0 m/s, u = 29.0 m/s, S = 1.20 m,
Substituting these values into equation 2
t = 2(1.2)/(29+0)
t = 2.4/29
t = 0.83 sec.
<em>Thus the collision last 0.83 s</em>
Atom is a piece of matter and matter makes up everything in the universe. I think this is the answer