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labwork [276]
3 years ago
9

A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s

. Determine the wheel's average rotational acceleration. Express your answer in radians per second squared. Enter positive value if the rotational speed increases and negative value if the rotational speed decreases.

Physics
2 answers:
Likurg_2 [28]3 years ago
7 0

Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)

Explanation: Please see the attachments below

sesenic [268]3 years ago
5 0

Answer:

The average rotational acceleration is -0.4 rad/s²

Explanation:

Given:

w₁ = initial rotational speed = 4 rad/s

w₂ = final rotational speed = 2 rad/s

t = time = 5 s

The average rotational acceleration is:

w_{average} =\frac{w_{2}-w_{1}  }{t} =\frac{2-4}{5} =-0.4rad/s^{2}

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A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T
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Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

m_pv_p=v_c(m_p+m_b) where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

0.5(m_p+m_b)v_c^{2}=0.5kx^{2} where k is spring constant and x is the compression of spring. Substituting the given values then

(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m

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A long solenoid has a radius of 2.0 cm and has 700 turns/m. If the current in the solenoid is decreasing at the rate of 8.0 A/s,
Brrunno [24]

Answer:

The magnitude of the induced electric field at a point 2.5 cm from the axis of the solenoid is 8.8 x 10⁻⁵ V/m

Explanation:

given information:

radius, r = 2.0 cm

N = 700 turns/m

decreasing rate, dI/dt = 9.0 A/s

the magnitude of the induced electric field at a point 2.5 cm (r = 2.5 cm = 0.025 m) from the axis of the solenoid?

the magnetic field at the center of solenoid

B = μ₀nI

where

B = magnetic field (T)

μ₀ = permeability (1.26× 10⁻⁶ T.m/A)

n = the number turn per unit length (turn/m)

I = current (A)

dB/dt = μ₀n dI/dt                                           (1)

now we calculate the induced electric field by using

E = \frac{1}{2}r\frac{dB}{dt}  

\frac{dB}{dt} = 2E/r                                                     (2)

where

E = the induced electric field (V/m)

we substitute the firs and second equation, thus

dB/dt = μ₀n dI/dt  

2E/r = μ₀n dI/dt  

E = (1/2) r μ₀n dI/dt

  = (1/2) (0.025) (1.26× 10⁻⁶) (700) (8)

  = 8.8 x 10⁻⁵ V/m

6 0
3 years ago
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