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Kipish [7]
3 years ago
12

Three wheels each of radius 1 have their centers at respective verticies of an equilateral triangle of side length 4. A belt is

wrapped continuously around the wheels. Find the length of the belt.
Physics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

L  = 6 * ( π + 1 )

Explanation:

The side of the equilateral triangle is  4, and each one of the circles is of radius 1. Then

Triangle vertex   A  B  and C

Trajectory of the belt, beginning in vertex A

1.-First circle  A  one turn

L₁  = 2*π*1   = 2*π

2.-Length between the circle A and B

L₂  = 2

3.- To wrap this circle we need to wrap the circle and to run the belt through the radius twice. The firs over the side AB and the second over the side BC, therefore

L₃ = 2*π*1 + 2*(1)

L₃ = 2*π + 2

4.-Length between two nearest points of circles B and C is 2 and length of the circle in C is 2*π*1. Then

L₄  = 2 + 2*π

Total length of the belt  is:

L  =  L₁ + L₂ + L₃ + L₄

L = 2*π + 2 + (2*π + 2 ) + ( 2 + 2*π )

L  = 6*π + 6

L  = 6 * ( π + 1 )

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When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. If the distance to the
Serjik [45]

Answer:

a) 5.61 rad/s

b) 16.83 rad/s

Explanation:

Distance to the floor = 79cm = 0.79m

Rotation is <1 revolution

The toast will rotate at an angular speed that is constant while it falls. The toast will also be falling with constant acceleration due to gravity. Using equation of motion, which is

S = ut + 1/2gt²

S = 0 + 1/2gt²

S = gt², where, S = d

0.79 = 9.81t²

t² = 0.79/9.81

t² = 0.081

t = 0.28s

As the toast is accidentally pushed, it rotates as it falls. It will be landing on its edges when and if it hits the ground. The smallest angle here would then be 1/4 of the revolution. This is also the smallest angular speed.

ω(min) = ΔΦ revolution /Δt

ω(min) = 1/4 * 2π / 0.28

ω(min) = 0.5π/0.28

ω(min) = 5.61 rad/s

Since 1/4 of revolution is the minimum angle, the remaining(3/4), is the maximum angle. Thus

ω(max) = 3/4 * 2π / 0.28

ω(max) = 0.75 * 2π / 0.28

ω(max) = 16.83 rad/s

8 0
4 years ago
Read 2 more answers
The 10-lb block A attains a velocity of 2ft/s in 5 seconds, starting from rest. Determine the tension in the cord and the coeffi
Nezavi [6.7K]

Answer:

Explanation:

Let T be the tension in the cord.

Impulse by cord = change in momentum of block A .

T x 5s = 10 ( 2 -0) = 20

T = 4 poundal .

acceleration of block B = 2 / 5 = 0.4 m /s²

Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .

= 8 ( 32 + .4 ) = 259.2 poundal

Frictional force on block A = 259.2 - 4 = 255.2 poundal

μ x 10 x 32 = 255.2

320μ = 255.2

μ =0 .8 .

8 0
3 years ago
an aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. the hole is 30mm in diameter and i
zavuch27 [327]

Answer:

ΔL = 1.011 mm

Explanation:

Let's begin by listing out the given information:

Length (L) = 600 mm = 0.6 m,

Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,

Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,

Modulus of Elasticity (E) = 85 GN/m²,

Compressive load (F) = 180 KN

Using the formula, Stress = Load ÷ Area

Mathematically,

σ = F ÷ A = 180 x 10³ ÷ 0.001256

σ = <u>143312.1 KN/m</u>²

Modulus of elasticity = stress ÷ strain

E = σ ÷ ε

ε = ΔL/L

85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)

ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶

ΔL = L x 1686.02 * 10⁻⁶

ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶

ΔL = 1.011 x 10⁻³ m

ΔL = <u>1.011 mm</u>

∴The bar contracts by 1.011 mm

4 0
3 years ago
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
just olya [345]

Answer:

2.78\times 10^{-35}\ \text{kg m/s}

6.178\times 10^{-34}\ \text{m/s}

0.31\times 10^{-4}\ \text{m/s}

Explanation:

\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

m = Mass of object

From Heisenberg's uncertainty principle we know

\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

The minimum uncertainty in the momentum of the object is 2.78\times 10^{-35}\ \text{kg m/s}

Golf ball minimum uncertainty in the momentum of the object

m=0.045\ \text{kg}

Uncertainty in velocity is given by

\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

The minimum uncertainty in the object's velocity is 6.178\times 10^{-34}\ \text{m/s}

Electron

m=9.11\times 10^{-31}\ \text{kg}

\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}

The minimum uncertainty in the object's velocity is 0.31\times 10^{-4}\ \text{m/s}.

6 0
3 years ago
nan moved 18m to the right and then another 22m to the right if the motion takes 20 seconds what is nans velocity
IrinaVladis [17]
Step 1: list known info

distance(change in position (Δx))= 18m+22m= 40m
time= 20 seconds

Step 2 :solve for velocity

velocity= Δx÷time
v= 40/20= 2m/s

Answer: the velocity is 2 meters per a second (m/s)


7 0
4 years ago
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