A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t
1 answer:
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Answer:
The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.
Explanation:
The average speed of gas molecules is given by:
R is the gas constant, T is the temperature and M the molar mass of the gas.
We know that a water molecule has a mass that is 18 times that of a hydrogen atom:
So, we have:
The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:
Answer:
The correct option is;
c. 22.6
Explanation:
The given parameters are;
The hypotenuse of the vector = 32
The angle of the vector = 45°
Therefore, the vector component in the y-axis is given as follows;
Substituting the values from the question gives;
The vector component in the y-axis, , is approximately 22.6.
By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.
ANGULAR MOTION
Since the car accelerate from rest, initial velocity will be zero.
Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters
- Initial velocity U = 0
- Final velocity V = 24.3 m/s
- Time t = 9.1 s
If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.
First equation
V = U + at
Substitute all necessary parameters into the equation.
24.3 = 0 + 9.1a
a = 24.3/9.1
a = 2.67 m/
Third Equation of motion
Substitute all the necessary parameters
= 0 + 2 x 2.67 x S
590.49 = 5.34S
S = 590.49 / 5.34
S = 110.58 m.
Given that the diameter of a tire is 80.2 cm,
the radius (r) will be 80.2/2 = 40.1 cm
convert it to meter
r = 40.1/100 = 0.401 m
The Circumference of the tire = 2r
Circumference = 2 x 3.143 x 0.401
Circumference = 2.52 m
Assuming no slipping, number of revolutions = 110.58/2.52
Number of revolutions = 43.89 rev.
Number of revolutions = 43 rev.
Therefore, the number of revolutions the tire makes during this motion is 43 rev.
Learn more about circular motion here: brainly.com/question/6860269