Answer:
a). Determine the magnitude of the gravitational force exerted on each by the earth.
Rock:
Pebble:
(b)Calculate the magnitude of the acceleration of each object when released.
Rock:
Pebble:
Explanation:
The universal law of gravitation is defined as:
(1)
Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.
<em>Case for the rock </em><em>:</em>
m1 will be equal to the mass of the Earth and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth .
Newton's second law can be used to know the acceleration.
(2)
<em>Case for the pebble </em><em>:</em>
Explanation:
Mass of the diskshaped grindstone, m = 1.1 kg
Radius of disk, r = 0.09 m
Angular velocity,
Time, t = 42.4 s
We need to find the frictional torque exerted on the grindstone. Torque in the rotational kinematics is given by :
I is moment of inertia of disk,
So, the frictional torque exerted on the grindstone is .
Answer:
38.8 km/hr
Explanation:
The total distance is 2 × 282 km = 564 km.
The total time is 6 hr 32 min + 8 hr = 14 hr 32 min = 14.533 hr.
The average speed is:
(564 km) / (14.533 hr)
38.8 km/hr
Answer:
11460 years
Explanation:
Let N₀ be the initial amount of carbon present. After one half-life, its value is N₀/2, after two half-lives, it is N₀/4 = N₀/2². After n nalf-lives, it is N = N₀/2ⁿ.
Now, if the value of carbon-14 drops to a quarter of its initial value, then N = N₀/4.
So, N₀/4 = N₀/2ⁿ
1/4 = 1/2ⁿ
1/2² = 1/2ⁿ
Comparing the exponents, n = 2.
So the value of carbon-14 present drops to a quarter of its initial value in 2 half-lives.
Since one-half life equals 5730 years, then two-half-lives is 2 × 5730 years = 11460 years
So, it takes carbon-14 11460 years for the number of nuclei to drop to a quarter of its initial value.