Answer:
the car with the hay should slow to 16m/s if the bale of hay is dropped into it.
Answer:
α = 395 rad/s²
Explanation:
Main features of uniformly accelerated circular motion
A body performs a uniformly accelerated circular motion when its trajectory is a circle and its angular acceleration is constant (α = cte). In it the velocity vector is tangent at each point to the trajectory and, in addition, its magnitude varies uniformly.
There is tangential acceleration (at) and is constant.
at = α*R Formula (1)
where
α is the angular acceleration
R is the radius of the circular path
There is normal or centripetal acceleration that determines the change in direction of the velocity vector.
Data
R = 0.0600 m :blade radius
at = 23.7 m/s² : tangential acceleration of the blades
Angular acceleration of the blades (α)
We replace data in the formula (1)
at = α*R
23.7 = α*(0.06)
α = (23.7) / (0.06)
α = 395 rad/s²
It would move to the right because the force is being applied from the left.
Answer:
These atomos are called isotopes.
Explanation:
Each chemical element is characterized by the number of protons in its nucleus, which is called the atomic number (Z).
The number of neutrons in the nucleus can vary. There are almost always as many or more neutrons than protons. The atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus.
The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element. That is, isotopes are atoms whose atomic nuclei have the same number of protons but different numbers of neutrons.
So, <u><em>these atomos are called isotopes.</em></u>
At 60 mph (26.7 m/s), Johnny Hotfoot slams on the brakes of his car and skids to a stop in 4 s. (A) How fast is the car slowing down? (b) The extent of the skid.
If the car had been traveling at 40.0 m/s, it would have slid 27.2 m.
The frictional force opposes the linear force pulling on the automobile. Consequently, m is the vehicle's mass, an is its acceleration, and c is its coefficient of friction.
The normal force is R.
given the aforementioned variables
Find the length of the skid if the vehicle had been traveling at 40.0 m/s.
The car would have skidded 27.2 meters if it had been driving at 40.0 meters per second.
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