a. 7.0 m/s
First of all, we need to convert the angular speed (1200 rpm) from rpm to rad/s:
Now we know that the row is located 5.6 cm from the centre of the disc:
r = 5.6 cm = 0.056 m
So we can find the tangential speed of the row as the product between the angular speed and the distance of the row from the centre of the circle:
b. 
The acceleration of the row of data (centripetal acceleration) is given by
where we have
v = 7.0 m/s is the tangential speed
r = 0.056 m is the distance of the row from the centre of the trajectory
Substituting numbers into the formula, we find
Use the inverse square law, thus if you move a distance of 3m away, the sound intensity decrease by 1/3^2= 1/9
Answer:
1.129×10⁻⁵ N
1.295 m
Explanation:
Take right to be positive. Sum of forces on the 31.8 kg mass:
∑F = GM₁m / r₁² − GM₂m / r₂²
∑F = G (M₁ − M₂) m / r²
∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²
∑F = 1.129×10⁻⁵ N
Repeating the same steps, but this time ∑F = 0 and we're solving for r.
∑F = GM₁m / r₁² − GM₂m / r₂²
0 = GM₁m / r₁² − GM₂m / r₂²
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
516 / r² = 207 / (0.482 − r)²
516 (0.482 − r)² = 207 r²
516 (0.232 − 0.964 r + r²) = 207 r²
119.9 − 497.4 r + 516 r² = 207 r²
119.9 − 497.4 r + 309 r² = 0
r = 0.295 or 1.315
r can't be greater than 0.482, so r = 0.295 m.
It’s burning wax. All of the other options are physical changes .
The car's mass is 1600 kg.
Its weight is (mass) x (gravity).
On Earth, that's (1600 kg) x (9.8 m/s²) = 15,680 Newtons.
At the moment, that's the only force acting on the car, directed downward and provided by gravity.
If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.
So the minimum force required to lift the car is <em>15,680 Newtons</em>, directed vertically upward.
