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.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
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?
Learn more about the Newton's equation of motion here:
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Force = (mass) x (acceleration)
= (0.025 kg) x (5 m/s²)
= 0.125 Newton
Answer:
<em>1.228 x </em>
<em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x 
N
modulus of elasticity E = 85 GN/m^2 = 85 x 
Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A = 
= 
= 1256.8 mm^2
area of hole a = 
= 
= 549.85 mm^2
Total contraction of the bar = 
total contraction = 
==> 
= <em>1.228 x </em><em> mm </em>
Answer:
time taken by the wave to reach the person is 0.2 s
Explanation:
As we know that the speed of the wave is given as
here we know that the wavelength of the wave is
now speed of the wave is given as
Now time taken by the wave to reach 5 m distance is
Unscrambling
1. resting heart rate
2. overload
3. workout
4. specificity
5. cool-down
6. progression
7. warm-up
8. the last one can only be instance, but there was a typo on the paper.
