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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Velocity is define as how fast an object is moving, and in what direction, it is a vector quantity, meaning velocity has both magnitude and direction. Anything goes to the left is negative, and anything goes to the right is positive.
a. Direction from east to west, given distance 11.5 meters, and time of 7.10 s
V = displacement/time V = -11.5/7.10 S V = -1.62 m/s (going left)
b. Joaquin reaches his original position. Displacement is now zero.
Velocity of the lawnmower is equal to "zero" but if we calculate for the average speed of the lawn, you just have to add the distance covered and the time it take to go back at the original position or point of origin
Answer:
The frequency of the photon is 
.
Explanation:
Given that,
Energy 
We need to calculate the energy
Using relation of energy
Where, 
= energy spacing
Put the value of h into the formula
Hence, The frequency of the photon is .
