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Ne4ueva [31]
3 years ago
14

A wheel starts from rest and rotates with constant angular acceleration and reaches an anaular speed of 12 rad/s in 3 s. Find (a

) the magnitude of the angular acceleration of the wheel and (b) the angle (in radian) through which it rotates in this time.
Physics
1 answer:
Vadim26 [7]3 years ago
8 0

Answer:

(a) 4 rad/s^2

(b) 18 rad

Explanation:

w0 = 0, w = 12 rad/s, t = 3 s

(a) Let α be the angular acceleration.

w = w0 + α t

12 = 0 + 3 α

α = 4 rad/s^2

(b) Let θ be the angle rotated

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 4 x 9

θ = 18 rad

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Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k
Harrizon [31]

Explanation:

Given that,

(a) Speed, v=6.66\times 10^6\ m/s

Mass of the electron, m_e=9.11\times 10^{-31}\ kg

Mass of the proton, m_p=1.67\times 10^{-27}\ kg

The wavelength of the electron is given by :

\lambda_e=\dfrac{h}{m_ev}

\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}

\lambda_e=1.09\times 10^{-10}\ m

The wavelength of the proton is given by :

\lambda_p=\dfrac{h}{m_p v}

\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}

\lambda_p=5.96\times 10^{-14}\ m

(b) Kinetic energy, K=1.71\times 10^{-15}\ J

The relation between the kinetic energy and the wavelength is given by :

\lambda_e=\dfrac{h}{\sqrt{2m_eK}}

\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}

\lambda_e=1.18\times 10^{-11}\ m

\lambda_p=\dfrac{h}{\sqrt{2m_pK}}

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\lambda_p=2.77\times 10^{-13}\ m

Hence, this is the required solution.

6 0
3 years ago
A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor
Dahasolnce [82]

Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is  located on the x-axis=\lambda_1

Linear charge density of another long line which is  located on the y-axis=\lambda_2

We have to find the direction of electric field at z=a on the positive z-axis if \lambda_1 and \lambda_2 are positive.

The direction of electric field  at z=a on the positive z-axis  is positive z-direction .

Because \lambda_1 and \lambda_2 are positive and the electric field is  applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

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pogonyaev

Explanation:

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Answer:

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