The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
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Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
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Answer:
Explanation:
The rotation rate of the man is:
The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:
![(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega](https://tex.z-dn.net/?f=%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%5Ccdot%20%280.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20%3D%20%5B%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%2B20000%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%5D%5Ccdot%20%5Comega)
The final angular speed is:
(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in
(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s
(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:
(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s
(c) The plane accelerates toward the center of the path with magnitude
<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²
(d) By Newton's second law, the tension in the line is
<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N
I would choose B. exosphere. Astronomers study outer space, planets and ect.
False because compound means two so a chemical compound is made up of two chemicals
