In electronics, the SI unit for current is Ampere. It is the amount of charge in Coulombs per unit time. It is named after the father of electrodynamics, Andre-Marie Ampere. Also, the current can be easily determined through the Ohm's Law, which states that current is equal to volts divided by the resistance. The answer is letter D.
<span>If Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second, and It hits stationary ball B and they undergo elastic collision, thus the two balls have different masses, then the following statement which is true is the statement that stated that there was no y-momentum initially.</span>
The answer is B because the pollinators give pollen from the plant
Answer:
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Explanation:
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?
It decreases in speed on its way down and increases in speed on its way down.
it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center
.It increases in speed on his way down because its under the influence of gravity
from newton's equation of motion we can check by
using V^2=u^2+2as
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Answer:
The energy dissipated as the puck slides over the rough patch is 1.355 J
Explanation:
Given;
mass of the hockey puck, m = 0.159 kg
initial speed of the puck, u = 4.75 m/s
final speed of the puck, v = 2.35 m/s
The energy dissipated as the puck slides over the rough patch is given by;
ΔE = ¹/₂m(v² - u²)
ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)
ΔE = -1.355 J
the lost energy is 1.355 J
Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J
