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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3^3 is the way you write 3³ when you don't have any way to write the ³ .
Whichever way you write it, it means " 3 · 3 · 3 " . That's 27 .
I don't know what would be considered "simplifying" it. Maybe " 3³ "is a little simpler, but you can't always write that if your computer or your word processor doesn't give you a way to write exponents.
" 3 · 3 · 3 " certainly is not any simpler than " 3^3 ", so that's not it.
" 27 " may be simpler.
In a spectrograph, black lines can be seen going through the array of colors. The pattern of these lines indicate the composition of the star.
Different elements block different parts of the spectrum, resulting in black lines.
Answer:
Explanation:
current I = 14 x 10⁻³ A
distance of current d = 2.2 x 10⁻² m
Magnetic field B due to a long straight current carrying wire
B = (μ₀ /4π) x (2 I / d )
= 10⁻⁷ x 2 x 14 x 10⁻³ / 2.2 x 10⁻²
= 12.72 x 10⁻⁸ T .