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2 years ago

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A square weightless frame ABCD with the side a is rotating around the side AB with the constant angular velocity ω1. A uniform d

isk of radius R and mass m is rotating around the frame’s diagonal AC with the constant angular velocity ω2. The center of the disk coincides with the center of the frame. Assuming ω1 = ω2 = ω find
a. The kinetic energy of the system.
b. The magnitude of the torque that needs to be applied to the axis AB to keep it (the axis AB) still.

1 answer:

Zarrin [17]2 years ago

5 0

Answer:

1) KE of the system = 1/2(Iw^2)

2) the torque T that must be applied to keep it still must be equal to the torque of rotation. This is equal to

T = (Iw^2)/2©

Where;

I = moment of inertia of the body

W = angular velocity of the body

© = angle of rotation in rad

Explanation:

Detailed explanation and calculation is shown in the image below

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A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprisi

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Answer:

v= 2413.5 m/s

Explanation:

maximum change of speed of rocket

=(initial exhaust velocity)×ln [(initialmass/finalmass)]

let initial mass= m

final mass = m-m(4/5) = m/5

[since the 80% of mass which is fuel is exhausted]

V-0 = 1500 ln (1/0.2)

V= 1500×1.609 = 2413.5 m/s

therefore, its exhaust speed v= 2413.5 m/s

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2 years ago

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An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

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2 years ago

A 13,000-N vehicle is to be lifted by a 25-cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter pis

MaRussiya [10]

Answer:

Explanation:

We shall apply Pascal's Law in fluid mechanics

According to it , pressure is transmitted in liquid from one point to another without any change .

25 cm diameter = 12.5 x 10⁻² m radius

Area = 3.14 x (12.5 x 10⁻²)²

= 490.625 x 10⁻⁴ m²

Pressure by vehicle

Force / area

13000 / 490.625 x 10⁻⁴

= 26.497 x 10⁴ Pa

5 cm diameter = 2.5 x 10⁻² radius

area = 3.14 x (2.5 x 10⁻²)²

= 19.625 x 10⁻⁴ m²

If we assume required force F on this area

Pressure = F / 19.625 x 10⁻⁴ Pa

According to Pascal Law

F / 19.625 x 10⁻⁴ = 26.497 x 10⁴

F = 19.625 x 26.497

= 520 N

4 0

2 years ago

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The

Eddi Din [679]

Answer:

(a) Steel rod: $1.1 * 10^{-4}$

Copper rod: $1.88 * 10^{-4}$

(b) Steel rod: $8.3 * 10^{-5} m$

Copper rod: $1.41 * 10^{-4} m$

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = $\frac{1}{Y} * \frac{F}{A}$

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y = 200 000 000 000 $N/m^{2}$

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

∴ $Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}$

For the copper rod:

Y = 120 000 000 000 N/m²

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

$Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain = = 1.88 * 10^{-4}$

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = $1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m$

For the copper rod:

Elongation = $1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m$

6 0

2 years ago