Answer:
The answer is 9.8 N
Explanation:
As we know that the weight of an object is the amount of gravitational force acting on the object in an upward direction if the weight is acting is the downward direction.
The formula of weight:
W = Mass x Gravitational force
W = m x g
Given data:
Mass =1 kg
g = 9.8 ms-2
W = 1kg x 9.8 ms-2 = 9.8 kgms-2 ( 1 kgms-2 = N)
SO,
W = 9.8 N
The toy has an earth weight of 9.8 N.
Answer:
3.036×10⁻¹⁰ N
Explanation:
From newton's law of universal gravitation,
F = Gm1m2/r² .............................. Equation 1
Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.
G = gravitational constant
Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²
Substituting into equation 1
F = 6.67×10⁻¹¹×7.9×6.1/2²
F = 321.427×10⁻¹¹/4
F = 30.36×10⁻¹¹
F = 3.036×10⁻¹⁰ N
Hence the force between the balls = 3.036×10⁻¹⁰ N
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.
v is the speed of cat, 
So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
Answer:
a 200 feet, and trains go a whole mile even after hitting the brakes
Explanation:
