Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
Answer:
ANSWER BELOW I
I
V
Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in
m/s2
. For example, for Earth, 445 N = 45.4 × 9.8
m/s2
:Notice that the x-axis values will be gravity in
m/s2
, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).
K=1400*V^2/2
K=20000*25^2/2. => 1400*V^2/2=20000*25^2/2 <=> 1400*V^2=20000*25^2
14*V^2=200*225
v^2=100*225/7
v=250/7^(1/2)
Answer: 250*7^(1/2)/7
The movement of electrical charges inside the earth
Answers:
kinetic energy lost = 86.4J
Explanation:
let Kf be the kinetic energy after the collision and Ki be the kinetic energy before the collision. let the 3kg car be 1 and 2kg car be 2.
Kf = K1(f) + K2(f)
Ki = K1(i) + k2(i)
loss in kinetic energy = Kf - Ki
= 1/2(3)(2.20)^2 + 1/2(2)(2.20)^2 - 1/2(3)(7)^2 - 1/2(2)(-5)^2
= 12.1 - 98.5
= -86.4 J
therefore, the kinetic energy lost in the collision is 86.4 J.
