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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D) decreasing the temperature lowers the average kinetic energy of the reactants.
Answer:
λ = 3.2 x 10⁻⁷ m = 320 nm
Explanation:
The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:
v = fλ
where,
v = c = speed of the electromagnetic waves (UV rays) = speed of light
c = 3 x 10⁸ m/s
f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz
λ = wavelength of the electromagnetic waves (UV rays) = ?
Therefore, substituting the values in the relation, we get:
3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)
λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)
<u>λ = 3.2 x 10⁻⁷ m = 320 nm</u>
So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.
Answer:
150 million kilometres
Explanation:
The astronomical unit (symbol: au, or AU or AU) is a unit of length, roughly the distance from Earth to the Sun and equal to 150 million kilometres (93 million miles) or 8.3 light minutes.
The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.
What does Gauss Law state?
It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.
Where,
= electric force = 4286 N/C
= Coulomb constant = 
= charges = ?
= distance of separation = 2.5 m
Put the values in the formula,
Therefore, the magnitude of the source charge is 3 μC.
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