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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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3 years ago
While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into a room an
Bogdan [553]

Answer:

the time needed for her to close the door is 1.36 s.

Explanation:

given information:

Force, F = 220 N

width, r = 1.40 m

weight, W = 790 N

height, h = 3.00 m

angle, θ = 90° = π/2

to find the times needed to close the door we can use the following equation

θ = ω₀t + 1/2 αt²

where

θ = angle

ω = angular velocity

α = angular acceleration

t = time

in this case, the angular velocity is zero. thus,

θ = 1/2 αt²

now, we can find the angular speed by using the torque formula

τ = I α

where

τ = torque

I = Inertia

we know that

τ = F r

and

I = 1/3 mr²

so,

τ = I α

F r = 1/3 mr² α

α = 3 F/mr

  = 3 F/(w/g)r

  = 3 (220)/(790/9.8) 1.4

  = 5.85 rad/s²

θ = 1/2 αt²

π/2 = 1/2 5.85 t²

t = 1.36 s

5 0
3 years ago
Question 5 of 10
Nina [5.8K]
It is D 14 MS. West. Kauehfbfnd
6 0
3 years ago
Bacteria which do not live in extreme conditions are found in which kingdom:
mel-nik [20]
Part of the
archeabactria
8 0
3 years ago
Two blocks of masses m1 and m2 are connected to each other on an Atwood machine (the blocks are connected by a string going over
ddd [48]

Answer:211.11 gm

Explanation:

Given

mass of lighter block m_1=100 gm

mass of heavier  block is m_2

acceleration of system a=3.5 m/s^2

From diagram

for lighter block

T-m_1g=m_1a

T=m_1(g+a)

For heavier Block

m_2g-T=m_2a

T=m_2(g-a)

Equating Tension T

m_1(g+a)=m_2(g-a)

m_2=m_1\cdot \frac{g+a}{g-a}

m_2=m_1\cdot \frac{9.8+3.5}{9.8-3.5}

m_2=0.1\cdot 2.11

m_2=0.2111 kg

m_2=211.11 gm

8 0
3 years ago
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