In a gear train with two gears, the gear ratio is defined as follows
where
is the angular velocity of the input gear while
is the angular velocity of the output gear.
This can be rewritten as a function of the number of teeth of the gears. In fact, the angular velocity of a gear is inversely proportional to the radius r of the gear:
But the radius is proportional to the number of teeth N of the gear. Therefore we can rewrite the gear ratio also as
Answer:
<h2>9.8 m/s²</h2>
Explanation:
<h2>Since the ball rises for 2.5 s, the time to fall is 2.5 s. The acceleration is 9.8 m/s2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s² downward.</h2>
Answer:
21 psi
Explanation:
The weight of the car is:
W = mg
W = 1000 kg * 9.8 m/s²
W = 9800 N
Divided by 4 tires, each tire supports:
F = W/4
F = 9800 N / 4
F = 2450 N
Pressure is force divided by area, so:
P = F / A
P = (2450 N) / (0.13 m × 0.13 m)
P ≈ 145,000 Pa
101,325 Pa is the same as 14.7 psi, so:
P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)
P ≈ 21 psi
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,
The circumference of the planet is,
Under the mathematical value the radius would be
Using second equation of motion
Replacing the values given,
Rearranging and solving for 'a' we have,
Using the value of acceleration due to gravity from Newton's law we have that
Here,
r = Radius of the planet
G = Gravitational Universal constant
M = Mass of the Planet
Therefore the mass of this planet is 
