The equation
(option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:


Where:
- m₁: is the mass of the lab cart = 15 kg
- m₂: is the <em>mass </em>of the object dropped = 2 kg
: is the initial velocity of the<em> lab cart </em>
: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

Therefore, the equation
represents the horizontal momentum (option 3).
Learn more about linear momentum here:
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Answer: 117.6N
Explanation:
By the second Newton's law, we know that:
F = m*a
F = force
m = mass
a = acceleration
We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.
Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:
F = 12kg*9.8m/s^2 = 117.6N
Answer:
<h2>Derived quantities are based on fundamental quantities, and they can be given in terms of fundamental quantities.</h2>
<h3>Fundamental quantities are the base quantities of a unit system, and they are defined independent of the other quantities. </h3>
Explanation:
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Answer:
The correct answer is 231 Mpa i.e option a.
Explanation:
using the equation of torsion we Have

where,
= shear stress at a distance 'r' from the center
T = is the applied torque
= polar moment of inertia of the section
r = radial distance from the center
Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.
We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

Applying values we get
