What is the acceleration of a car that goes from 4m/s to 8m/s in 2s use the guess method
1 answer:
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The true statement about the CD is:
<h3><em>b. No net torque acts on it at all.</em></h3>Centripetal Acceleration can be formulated as follows:
<em>a = Centripetal Acceleration ( m/s² )</em>
<em>v = Tangential Speed of Particle ( m/s )</em>
<em>R = Radius of Circular Motion ( m )</em>
Centripetal Force can be formulated as follows:
<em>F = Centripetal Force ( m/s² )</em>
<em>m = mass of Particle ( kg )</em>
<em>v = Tangential Speed of Particle ( m/s )</em>
<em>R = Radius of Circular Motion ( m )</em>
Let us now tackle the problem !
<em>Complete Question:</em>
<em>A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise. Which of the following statements about the CD is true?</em>
<em>a. A net torque acts on it clockwise to keep it moving</em>
<em>b. No net torque acts on it at all.</em>
<em>c. A net torque acts on it counterclockwise to keep it moving</em>
<u>Given:</u>
angular velocity = ω = 5.0 revolutions per second
<u>Asked:</u>
net torque = Στ = ?
<u>Solution:</u>
Constant angular velocity → angular acceleration = α = 0 rad/s²
The true statement about the CD is:
<em>b. No net torque acts on it at all.</em>
- Impacts of Gravity : brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
- The Acceleration Due To Gravity : brainly.com/question/4189441
Grade: High School
Subject: Physics
Chapter: Circular Motion
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
#LearnWithBrainly
Answer:
Option D) -0.0707
Explanation:
We are given the following in the question:
Population mean, μ = 45 mp
Sample mean, = 44.9
Sample size, n = 50
Sample standard deviation, s = 10
Formula:
Putting all the values, we have
Thus, the test statistic is
Option D) -0.0707
This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.
Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.
I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.
