Gravitational, gravitational ! both the option are same
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
Answer:
Intercellular communication refers to the communication between cells. Membrane vesicle trafficking has an important role in intercellular communications in humans and animals, e.g., in synaptic transmission, hormone secretion via vesicular exocytosis.
The frequency of a wave is a measure of the number of waves that passes through a point per unit of time. It has SI units of s^-1. It is also equivalent to Hertz (Hz). We calculate the frequency of the wave described as follows:
Frequency = 5 waves / 2 s = 2.5 / s or 2.5 Hz
