The diameter of the sphere is 2cm.
<h3>How to calculate the diameter?</h3>
From the diagram, the first sphere on the ruler is at 4cn and the last sphere is at 12cm.
Therefore, the length will be:
= 12 - 4.
= 8cm
The diameter of one sphere will be:
= Length / 4
= 8/4
= 2
Therefore, the diameter of the sphere is 2cm.
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That was sun as some smaller masses formed planets and other remaining formed sun
Answer is 6 tires.
This is a projectile question.
First make sure units are consistent - express speed in m/s.
20 km/h = 20000m / 3600 s = 5.56 m/s
Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.
Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:
Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s
An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:
Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s
Total trip time:
0.19 x 2 = 0.38s
Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:
v = d / t
5.22 = d / 0.38
d = 1.98m
Now divide this total distance by the length of an individual tire to find the number of tires he will clear:
1.98 / 0.3 = 6.6 tires
Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).
Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:
Where,
x= Displacement
= Initial velocity
a = Acceleration
t = time
Since there is no initial velocity, the same equation can be transformed in terms of length and time as:
For the second cart
When the tenth car is aligned the length will be 9 times the initial therefore:
When the tenth car has passed the length will be 10 times the initial therefore:
The difference in time taken from the second car to pass it is 5 seconds, therefore:
From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:
From the relationship when the car has passed and the time difference we will have to:
Replacing the value found in the equation given for the second car equation we have to:
Finally we will have the time when the cars are aligned is
The time when you have passed it would be:
The difference between the two times would be:
Therefore the correct answer is C.