Answer:
option B
Explanation:
given,
height of building = 0.1 km
ball strikes horizontally to ground at = 65 m
speed at which the ball strike = ?
vertical velocity = 0 m/s
time at which the ball strike
t = 4.53 s
vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s
horizontal velocity = 
=14.35 m/s
speed of the ball = 
= 46.65 m/s
hence, the speed of the ball just before it strike the ground = 47 m/s
The correct answer is option B
Given Information:
KEa = 9520 eV
KEb = 7060 eV
Electric potential = Va = -55 V
Electric potential = Vb = +27 V
Required Information:
Charge of the particle = q = ?
Answer:
Charge of the particle = +4.8x10⁻¹⁸ C
Explanation:
From the law of conservation of energy, we have
ΔKE = -qΔV
KEb - KEa = -q(Vb - Va)
-q = KEb - KEa/Vb - Va
-q = 7060 - 9520/27 - (-55)
-q = 7060 - 9520/27 + 55
-q = -2460/82
minus sign cancels out
q = 2460/82
Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹
q = 2460(1.60x10⁻¹⁹)/82
q = +4.8x10⁻¹⁸ C
Please find attached photograph for your answer.
Hope it helps.
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Answer: Solor cycles and tings
