# A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?

The value of final angular speed of the uniform rod which rests on the frictionless horizontal surface is,

### What is angular speed of a body?

The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it.

The mass of the bullet is one-fourth the mass of the rod. The diagram for the above condition is attached below.

In the attached image the angular momentum about the point A is constant just before and after the collision. Thus,

Put the value of inertia as,

Solving it further we get,

Hence, the value of final angular speed of the uniform rodwhich rests on the frictionless horizontal surface is,

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Explanation:

From Conservation of angular momentum, we know that

L1 = L2 ,

Therefore MV L/2 = ( Irod + Ib) x W

M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W

M/8 X VL = (ML^2/16 + ML^2 /3 )

After elimination we have,

V/8 = 19/48 x L x W

W = 48/8 x V/19L = 6/19 x V/L

Therefore W = (0.136)X V/L

## Related Questions

The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe is 75 mg/L. If the BOD is destroyed through a first-order reaction with a rate constant equal to 0.05/day, what is the BOD concentration 50 km downstream? The velocity of the river is 15 km/day.

The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L

Explanation:

Let the initial concentration of the BOD = C₀

Concentration of BOD at any time or point = C

dC/dt = - KC

∫ dC/C = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = -kt + b (b = constant of integration)

At t = 0, C = C₀

In 1 = 0 + b

b = 0

In (C/C₀) = - kt

(C/C₀) = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C₀ = 75 mg/L

k = 0.05 /day

C = 75 e⁻⁰•⁰⁵ᵗ

So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day

We calculate how many days it takes the river to reach 50 km downstream

Velocity = (displacement/time)

15 = 50/t

t = 50/15 = 3.3333 days

So, we need the C that corresponds to t = 3.3333 days

C = 75 e⁻⁰•⁰⁵ᵗ

0.05 t = 0.05 × 3.333 = 0.167

C = 75 e⁻⁰•¹⁶⁷

C = 63.5 mg/L

The BOD concentration 50 km downstream from the wastewater treatment plant is approximately 15.865 mg/L.

### Explanation:

To calculate the BOD concentration 50 km downstream, we need to consider the rate of dilution due to the flow of the river and the first-order reaction that destroys BOD. The concentration of BOD downstream can be calculated using the equation C2 = C1 * exp(-k * d/v), where C1 is the initial concentration, k is the rate constant, d is the distance, and v is the velocity of the river.

Plugging in the given values, we have C2 = 75 * exp(-0.05 * 50/15), which gives us a BOD concentration of approximately 15.865 mg/L 50 km downstream from the wastewater treatment plant.

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An aluminum rod is 10.0 cm long and a steel rod is 80.0 cm long when both rods are at a temperature of 15°C. Both rods have the same diameter. The rods are now joined end-to-end to form a rod 90.0 cm long. If the temperature is now raised from 15°C to 90°C, what is the increase in the length of the joined rod? The coefficient of linear expansion of aluminum is 2.4 × 10-5 K-1 and that of steel is 1.2 × 10-5 K-1.

0.9 cm

Explanation:

The computation in the increase in the length of the joined rod is shown below:

As we know that

Increase in length = increase in the length of aluminum rod + increase in The length of steel rod

= 0.9 cm

We simply added the length of aluminium rod and length of steel rod so that the length of the joined rod could come and the same is to be considered

The increase in length of the joined rod when the temperature is raised from 15°C to 90°C is 0.090 cm.

### Explanation:

To determine the increase in length of the joined rod when the temperature is raised from 15°C to 90°C, we need to use the formula for linear expansion: AL = aLAT, where AL is the change in length, AT is the change in temperature, and a is the coefficient of linear expansion. First, we need to calculate the change in temperature for each rod: ΔT = 90°C - 15°C = 75°C. For the aluminum rod, using a linear expansion coefficient of 2.4 × 10-5 K-1 and a length of 10.0 cm, we can calculate the change in length using the formula: ALaluminum = (2.4 × 10-5 K-1)(10.0 cm)(75°C) = 0.018 cm. Similarly, for the steel rod, using a linear expansion coefficient of 1.2 × 10-5 K-1 and a length of 80.0 cm, we can calculate the change in length: ALsteel = (1.2 × 10-5 K-1)(80.0 cm)(75°C) = 0.072 cm. Since the rods are joined end-to-end, the total change in length of the joined rod is the sum of the individual changes: ΔL = ALaluminum + ALsteel = 0.018 cm + 0.072 cm = 0.090 cm. Therefore, the increase in the length of the joined rod is 0.090 cm.

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Three point mass particles are located in a plane: a. 3.77 kg located at the origin
b. 6.7106 kg at [(5.72 cm),(11.44 cm)]
c. 2.46181 kg at [(16.7024 cm),(0 cm)].

How far is the center of mass of the three particles from the origin? Answer in units of cm

The distance of the center of mass of the three particles from the origin is 6.1428 cm and 5.9316 cm.

### Calculation of the distance:

Since

m1 = 3.77 kg (0, 0 )

m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)

m3 = 2.46181 kg (16.7024 cm, 0 cm )

Now here we assume x and y be the coordinates with respect to the centre of mass.

So,

We know that

= 6.1428 cm

Now

= 5.9316 cm

Explanation:

m1 = 3.77 kg (0, 0 )

m2 = 6.7106 kg ( 5.72 cm, 11.44 cm)

m3 = 2.46181 kg (16.7024 cm, 0 cm )

Let x and y be the coordinates of centre of mass.

x = 6.1428 cm

y = 5.9316 cm

Exposure to what type of radiant energy is sensed by human skin as warmth? x-rays ultraviolet infrared gamma rays

infrared radiant energy is sensed by human skin as warmth. Hence option C is correct.

Radiation in physics is the emission or transmission of energy as waves, particles, or both, via space or a material medium.[1][2] This comprises:

ultrasonography, sound, and seismic waves (reliant on a physical transmission medium) are examples of acoustic radiation.

gravity radiation, which manifests as gravitational waves or ripples in spacetime's curvature

Depending on the energy of the emitted particles, radiation is frequently divided into ionising and non-ionizing categories. More than 10 eV is carried by ionising radiation, which is sufficient to ionise atoms, molecules.

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i think it is infared

You observe three carts moving to the right. Cart A moves to the right at nearly constant speed. Cart B moves to the right, gradually speeding up. Cart C moves to the right, gradually slowing down. Which cart or carts, if any, experience a net force to the right

Explanation:

Cart A is moving to the right with constant speed i.e. net acceleration is zero

because acceleration is change in velocity in given time

Cart B is moving towards right with gradually speed up so there is net acceleration which helps to increase the velocity s

This indicates the net force acting on the cart towards right

For cart C there is gradual slow down of cart which indicates cart is decelerating and a net force is acting towards which opposes its motion.

A sample that includes important subgroups that the researchers want to be able to generalize their results to is called