Answer: distance d = 4.73e10m
Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.
Using electric potential V formula:
V = kq / d
Where K = 9.05×10^9Nm^2/C
And e = 1.6×10^-19C
But you don't need to substitute it.
1090 V = 8.99e9N·m²/C² * 5740C /d
Make d the subject of formula
d = 4.73e10 m
Answer:
The velocity will be "76.8 m/s".
Explanation:
The given values are:
Acceleration,
a = 2.4 m/s²
Time,
t = 32 seconds
By equation of motion,
⇒
On substituting the values, we get
⇒
⇒
⇒
Answer:
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Answer:
(C) 40m/s
Explanation:
Given;
spring constant of the catapult, k = 10,000 N/m
compression of the spring, x = 0.5 m
mass of the launched object, m = 1.56 kg
Apply the principle of conservation of energy;
Elastic potential energy of the catapult = kinetic energy of the target launched.
¹/₂kx² = ¹/₂mv²
where;
v is the target's velocity as it leaves the catapult
kx² = mv²
v² = kx² / m
v² = (10000 x 0.5²) / (1.56)
v² = 1602.56
v = √1602.56
v = 40.03 m/s
v ≅ 40 m/s
Therefore, the target's velocity as it leaves the spring is 40 m/s