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3 years ago

7

A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned

when its angular speed reaches 72.0 rad/s

1 answer:

Neporo4naja [7]3 years ago

5 0

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, $\omega _i$ = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, $\omega_f$ = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

$\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\ -\ \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\ -\ (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad$

Therefore, the angle through which the wheel turned is 947.7 rad.

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Answer:

The total energy of the composite system is 7.8 J.

Explanation:

Given that,

Height = 0.15 m

Radius of circular arc = 0.27 m

Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.

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We need to calculate the total energy of the composite system

Total energy of the system at any time = Potential energy of the system at the stopping point

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$E=(m+M)gh$

Put the value in to the formula

$E=(30\times10^{-3}+5.29)\times 9.8\times0.15$

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4 years ago

The diagram below shows the velocity vectors for two cars that are moving relative to each other.

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Answer:

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Explanation:

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Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

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Answer:

Part a)

$I = 1.5 kg m^2$

Part b)

$I = 0.75 kg m^2$

Part c)

$I = 1.5 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

$I = mL^2 + mL^2 + m(0) + m(0)$

$I = 2mL^2$

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$I = 1.5 kg m^2$

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

$I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2$

$I = 2m\frac{L^2}{2}$

$I = (3 kg)(0.50^2)$

$I = 0.75 kg m^2$

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

$I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2$

$I = 4m\frac{L^2}{2}$

$I = 2(3 kg)(0.50^2)$

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8 0

3 years ago