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.
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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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Because the alpha particles were passing through the foil with slight to no reflection however some alpha particles were deflected with high angle or reversed direction which is not what was expected as according to thomson the atom is a lump of postively charged mass and there will be no deflection as the charge is distrbuted and the charge of alpha particle is 207 times that of electron
Given Information:
Charge = q = 7.0x10⁻⁶ C
Required Information:
Number of electrons = n = ?
Answer:
Number of electrons = 4.375x10¹³
Explanation:
We know that charge is given by
q = n*e
Where n is the number of electrons and e is the electronic charge that is 1.60x10⁻¹⁹ C
n = q/e
n = 7.0x10⁻⁶/1.60x10⁻¹⁹
n = 4.375x10¹³ electrons
Therefore, 4.375x10¹³ electrons were transferred.
Answer:
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Explanation:
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