An object traveling 200 feet per second slows to 50 feet per second in 5 seconds calculate the object

A uniform rod is 2. 0 m long and has mass 15 kg. What is most nearly the rod's mass moment of inertia?

trapecia [35]

The rod's mass moment of inertia is 5kgm².

<h3>Moment of Inertia:</h3>

The "sum of the product of mass" of each particle with the "square of its distance from the axis of rotation" is the formula for the moment of inertia.

The Parallel axis Theorem can be used to compute the moment of inertia about the end of the rod directly or to derive it from the center of mass expression. I = kg m². We can use the equation for I of a cylinder around its end if the thickness is not insignificant.

If we look at the rod we can assume that it is uniform. Therefore the linear density will remain constant and we have;

or = M / L = dm / dl

dm = (M / L) dl

$I = \int\limits^M_0 {r^2} \, dm$

$I = \int\limits^\frac{L}{2} _\frac{-L}{2} {I^2 (M/L)} \, dl$

Here the variable of the integration is the length (dl). The limits have changed from M to the required fraction of L.

$I = \int\limits^\frac{L}{2} _\frac{-L}{2} {I^2 (M/L)} \, dl$

$I = \frac{M}3L}[(\frac{L^3}{2^3} - \frac{-L^3}{2^3} )]\\\\I = \frac{1}{12}ML^2$

Mass of the rod = 15 kg

Length of the rod = 2.0 m

Moment of Inertia, I = $\frac{1}{12}15 (2)^2$

= 5 kgm²

Therefore, the moment of inertia is 5kgm².

Learn more about moment of inertia here:

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4 0

2 years ago

A force F₁ of magnitude 6.50 units acts at the origin in a direction 49.0° above the positive x axis. A second force F₂ of magni

Sergeu [11.5K]

Found this hope it helps

4 0

4 years ago

What name is given to the property by which objects resist changes in motion?

Reptile [31]

Inertia is the tendency of an object to resist changes in its state of motion.

8 0

3 years ago

Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c

PIT_PIT [208]

Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
$E=k_e \frac{q}{r^2}$
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
$E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C$

While the electric field generated by the second charge is
$E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C$

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
$E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C$