Answer: 30
Explanation: it is 30 meters per seconds
First convert from mi/h to ft/s. There are 5280 ft to 1 mi, and 3600 s to 1 h, so
36 mi/h = (36 mi/h) * (5280 ft/mi) * (1/3600 h/s) = 52.8 ft/s
Let <em>a</em> be the acceleration of the car. The car's speed at time <em>t</em> is
<em>v</em> = 52.8 ft/s + <em>a</em> <em>t</em>
so that after 5.4 s, it attains a speed of
<em>v</em> = 52.8 ft/s + (5.4 s) <em>a</em>
Recall that
<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>x</em>
where <em>u</em> is the car's initial velocity and ∆<em>x</em> is the distance it's traveled.
We have
(52.8 ft/s + (5.4 s) <em>a</em>)² - (52.8 ft/s)² = 2 <em>a</em> (595 ft)
Omitting units, this equation reduces to
(52.8 + 5.4 <em>a</em>)² - 52.8² = 1190 <em>a</em>
==> 29.16 <em>a</em>² - 619.76 <em>a</em> = 0
==> 29.16 <em>a</em> - 619.76 = 0
==> 29.16 <em>a</em> = 619.76
==> <em>a</em> ≈ 21.25 ft/s²
Answer:
Explanation:
From frequency of oscillation
Initially with the suspended string, the above equation is correct for the relation, hence
where k is force constant and m is the mass
When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional
Employing f2/ f1, we have
Answer:
Explanation:
A and B are touched .
charge on each of them after touching = (9q - q) / 2 = 4q
when C is touched with A
charge on A and C each after touching
= 4q + 0 / 2 = 2q
When C is touched with B
charge on each of them
(2q + 4q ) / 2 = 3 q
Final charge on C = 3q
