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Anarel [89]
1 year ago
9

two identical, thin rods, each with mass mm and length ll, are joined at right angles to form an l-shaped object. this object is

balanced on top of a sharp edge (figure 1). if the l-shaped object is deflected slightly, it oscillates.
Physics
1 answer:
Allushta [10]1 year ago
3 0

The length of time that an action or state persists. The given L-shaped object made of two rods oscillates at a frequency of 2π√√2L/3g.

The rod weighs m grams.

The length of a single rod is L.

If it is connecting at, as indicated by the query, then the moment of inertia of the combined system follows.

mL²/3+3L²/3=2mL²/3

Here, I denotes the rod's moment of inertia, d denotes the separation of the center of gravity from the fixed point, and T denotes the length of time the L-shaped object oscillates.

Length of pendulam

Lcos45=L*1/√2=L/√2

Caluclate the time period

T=2π√I/mgl

=2π√2*mL²*√2/3*2mgl

oscillates at a frequency = 2π√√2L/3g

Learn more about frequency here:

brainly.com/question/10254661

#SPJ4

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yan [13]
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5 0
3 years ago
Find the volume of the tire with dimensions
Vinvika [58]

The volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

<h3>What is volume?</h3>

Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

<h3>Volume of the tire</h3>

The volume of the tire is the measure of the product of area and thickness of the tire.

The volume of the tire is calculated as follows;

Radius of the tire = 0.5 x 26" = 13"

Volume of the tire = Area x thickness

Volume of the tire = πr² x h

where;

  • r is the radius of the tire
  • h is the thickness of the tube

Volume of the tire = π(13)² x (2.125)

Volume of the tire =  1,128.2 cubic inch

Thus, the volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

Learn more about Volume of tire here: brainly.com/question/1972490

#SPJ1

7 0
2 years ago
A wave travels with speed 200 m/s. Its wave number is 1.5 rad/m. What are its (a) wavelength and (b) frequency
Vladimir [108]

Answer:

(a)

Explanation:

5 0
3 years ago
3. A rocket is launched at an angle of 53 degrees above the 1 point
irina1246 [14]

Answer:

24,000 m

Explanation:

First find the rocket's final position and velocity during the first phase in the y direction.

Given:

v₀ = 75 sin 53° m/s

t = 25 s

a = 25 sin 53° m/s²

Find: Δy and v

Δy = v₀ t + ½ at²

Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²

Δy = 7736.8 m

v = at + v₀

v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)

v = 559.0 m/s

Next, find the final position of the rocket during the second phase (as a projectile).

Given:

v₀ = 559.0 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 15945.5 m

The total displacement is:

7736.8 m + 15945.5 m

23682.2 m

Rounded to two significant figures, the maximum altitude reached is 24,000 m.

3 0
3 years ago
At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc
amm1812

Explanation:

Given that,

Mass of the rocket, m = 2150 kg

At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

F=At^2

Here F = 888.93 N when t = 1.25 s

(c) We can find the value of A first as :

F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2

The value of A is 568.91\ N/s^2.

(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

J=\int\limits {F{\cdot} dt}

Limits will be from 2 s to 2+ 4 = 6 s

It implies :

J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns

(b) Impulse is also equal to the change in momentum as :

J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s

Hence, this is the required solution.

5 0
3 years ago
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