What is the factor involved in increasing an object’s inertia?

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

2 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me

never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

$I=\frac{Q}{t}$

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

$I=\frac{420 C}{4800 s}=0.0875 A$

2) Drift velocity of the electrons: $1.78\cdot 10^{-6} m/s$

The drift velocity of the electrons in the wire is given by:

$u = \frac{I}{nAq}$

where

I = 0.0875 A is the current

$n=5.8\cdot 10^{28}$ is the number of free electrons per cubic meter

A is the cross-sectional area

$q=1.6\cdot 10^{-19} C$ is the charge of one electron

The radius of the wire is

$r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m$

So the cross-sectional area is

$A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2$

So, the drift velocity is

$u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s$

4 0

2 years ago

An object of mass M is dropped near the surface of Earth such that the gravitational field provides a constant downward force on

marysya [2.9K]

Answer:

The answer is: c. It does not move

Explanation:

Because the gravitational force is characterized by being an internal force within the Earth-particle system, in this case, the object of mass M. And since in this system there is no external force in the system, it can be concluded that the center of mass of the system will not move.

6 0

2 years ago

A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the

egoroff_w [7]

The acceleration of the jet is $28.1 m/s^2$

Explanation:

Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0

v = 75 m/s

s = 100 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{75^2-0}{2(100)}=28.1 m/s^2$

Learn more about acceleration:

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#LearnwithBrainly

4 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0

2 years ago