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?
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Answer:
3.42N
Explanation:
*not too sure bc i left my physics notes at school so it might not be 100% accurate :p*
Use the equation: F = (GMm)/(r^2)
F = force of gravity
G = gravitational constant (6.7x10^-11)
M = mass1 (2.5x10^30kg)
m = mass2 (1kg)
r = radius (7000m)
Plug it in: F = ((6.7x10^-11)(2.5x10^30)(1)) / (7000^2)
F = (1.675x10^20) / (4.9x10^7)
F = 3.4183673x10^12
F = 3.42N
They are said to be directly related.
a) directly related.
This is Charles' Law.
Answer:
FALSE!!!
Explanation:
Business letters are formal and normally given to someone of higher importance.
