What is the radius of an automobile tire that turns with a period of 0.091 s and has a linear
1 answer:
Answer:
1) The radius of the tire is approximately 0.28966 meters
2) The centripetal force is the force that keeps a body moving on a circular path
Explanation:
1) The linear speed of the automobile tire = 20.0 m/s
The period with which the tire turns = 0.091 s
The period = The time it takes to make a complete turn
Therefore;
The number of turns in 1 second = 1/0.091 ≈ 10.989 turns
The distance covered with 10.989 turns, assuming no friction = 10.989 × The circumference of the tire
∴ The distance covered with 10.989 turns, assuming no friction = 10.989 × 2 × π × Radius of the tire
From the speed of the car, 20.0 m/s, we have;
The distance covered in 1 second = 20.0 meters
Therefore;
10.989 × 2 × π × Radius of the tire = 20.0 meters
Radius of the tire = (20.0 meters)/(10.289 × 2 × π) ≈ 0.28966 meters
The radius of the tire ≈ 0.28966 meters
2) The centripetal force is the force required to maintain the curved motion of an object, and having a direction towards the center of the rotary motion.
The centripetal force is given by the formula,
Where;
F = The centripetal force
m = The mass of the object
v = The linear velocity of the object
r = The radius of the rotational motion.
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Answer:
(A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Explanation:
Given that,
A nucleus undergoes a nuclear decay.
(A). In alpha decay,
We know that,
When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Fr.
(B). In beta-minus decay,
We know that,
When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Po.
(C). In beta-plus decay,
We know that,
When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Ne.
(D). In gamma decay,
We know that,
When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Tc.
Hence, (A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Answer:
70.6 mph
Explanation:
Car A mass= 1515 lb
Car B mass=1125 lb
Speed of car B is 46 miles/h
Distance before locking, d=19.5 ft
Coefficient of kinetic friction is 0.75
Initial momentum of car B=mv where m is mass and v is velocity in ft/s
46 mph*1.46667=67.4666668 ft/s
Initial momentum of car A is given by
where
is velocity of A
Taking East as positive and west as negative then the sum of initial momentum is
The common velocity is represented as hence after collision, the final momentum is
From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence
The acceleration of two cars
From kinematic equation
hence
Substituting the value of in equation