a) 32.3 N
The force of gravity (also called weight) on an object is given by
W = mg
where
m is the mass of the object
g is the acceleration of gravity
For the ball in the problem,
m = 3.3 kg
g = 9.8 m/s^2
Substituting, we find the force of gravity on the ball:
b) 48.3 N
The force applied
The ball is kicked with this force, so we can assume that the kick is horizontal.
This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:
And substituting
W = 32.3 N
Fapp = 36 N
We find
c) 
The ball's acceleration can be found by using Newton's second law, which states that
F = ma
where
F is the net force on an object
m is its mass
a is its acceleration
For the ball in this problem,
m = 3.3 kg
F = 48.3 N
Solving the equation for a, we find
Answer:
a) a = 4.57 m/s², b) a = 6.48 m / s²
, c) a = 1.42 m / s²,d) r = 82.3 m
Explanation:
The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is
a = v² / r
let's apply this precaution to our cases
a) let's calculate
a = 8²/14
a = 4.57 m/s²
b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s
let's reduce feet to meters
1 ft = 0.3048 m
r = 165 ft (0.3048 m / 1 ft) = 50.292 m
a = 18,055 2 / 50,292
a = 6.48 m / s²
c) we calculate
a = 1.25²2 / 1.1
a = 1.42 m / s²
d) we look for the radius
a = v² / r
r = v² / a
we reduce
v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22 ms
r = 22.22²/6
r = 82.3 m
e) the cenripeta acceleration is used to take the curves on the highway,
Used in centrifuges to separate compounds
It is used in the games of the park of atraccio
Used in CD players and computer hard drives
Answer: 10.2 kg if g = 9.8, 10 if g = 10.
Explanation:
Weight or the "force of gravity" on a person is simply defined by the equation: F = ma. In this case, the acceleration is g, which is 9.8 but can be rounded up to 10. Based on this, we have:
F = mg
100 = m*9.8
m = 10.2(or 10 if we set g to 10).
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution = 
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:
Plug in the given values and solve for 
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².
The answer is focal point.
The focal length is the distance from the lens (or mirror) to the focal point. The focal point is <span>the point at which rays of light converge.</span>
