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maxonik [38]
4 years ago
5

A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i

t has reached angular velocity after 2 complete revolutions. What is the angular acceleration of the carousel during this time
Physics
1 answer:
OLga [1]4 years ago
7 0

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

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. Boxes are sitting on a conveyor belt as the conveyor is turned on, moving the boxes toward the right. The belt reaches full sp
oksian1 [2.3K]

Answer:

The acceleration of the boxes is 1.5 ft/s²

The displacement of the boxes during the speed-up period is 0.1875 ft.

Explanation:

Hi there!

Let´s convert the 45 ft/min into ft/s:

45 ft/min ·  1 min/ 60 s = 0.75 ft/s

It takes the belt 0.5 s to reach this speed. Then, the acceleration of the boxes will be:

a = v/t

Where:

a = acceleration.

v = velocity.

t = time.

a = 0.75 ft/s / 0.5 s

a = 1.5 ft/s²

The acceleration of the boxes is 1.5 ft/s²

The equation of displacement is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the boxes at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since the origin of the frame of reference is located at the point where the boxes begin to move, x0 = 0. Since the boxes were initially at rest, v0 = 0. Then:

x = 1/2 · a · t²

x = 1/2 · 1.5 ft/s² · (0.5 s)²

x = 0. 1875 ft

The displacement of the boxes during the speed-up period is 0.1875 ft.

5 0
3 years ago
A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma
Alexxx [7]

Answer:

Explanation:

Given

Mass of big box is M and small box is m

Tension T will cause the boxes to accelerate

T=(M+m)a

where a=acceleration of the boxes

Now smaller box will slip over large box if the acceleration force will exceed the static friction

i.e. for limiting value

\mu _smg=ma

a=\mu _s\cdot g

thus maximum tension

T=\mu _s(M+m)g

6 0
4 years ago
Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kg when weighed in air. The density of
sasho [114]

Answer:

m = 4.5021 kg

Explanation:

given,

Apparent mass of aluminium = 4.5 kg

density of air = 1.29 kg/m³

density of aluminium = 2.7 x 10⁷ kg/m³

true mass of the aluminium = ?

Weight in Vacuum

W = m g

W = ρV g

Air buoyancy acting on aluminium

B = ρ₀V g

Volume is the same in both cases since the volume of the aluminum

displaces an equal amount of volume air.

Apparent weight:

ρV g − ρ₀V g = 4.5 g

ρV − ρ₀V = 4.5

V = \dfrac{4.5}{\rho - \rho_0}

m = ρV

m = \dfrac{4.5\times \rho}{\rho - \rho_0}

m = \dfrac{4.5\times 2700}{2700 - 1.29}

m = 4.5021 kg

4 0
3 years ago
45 joules were expended to move a box weighing 30 newtons. How many meters was it moved?
slava [35]

Answer:

1.5 meter is the answer

7 0
3 years ago
PLEASE HELPPPPPP <333​
True [87]

Answer:

the answer is b

Explanation:

gravity pulls you down so on a scale you will weigh more. less gravity you will weigh less.

8 0
3 years ago
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