Answer:
both
Explanation:
and I'm serious he is literally both
Answer:
emf = 11.667 V
Explanation:
Given: charge q = 0.060 C, electric potential energy E =0.70 J,
Solution :
by definition 1 volt = 1 joule per coulomb
so Voltage = emf = E/C
emf = 0.70 J / 0.060 C
emf = 11.667 V
Answer:
the answer is 52.15cm×35.5
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
