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Fiesta28 [93]
4 years ago
12

A bicycle tire is spinning counterclockwise at 3.00 rad/s. During a time period Δt = 2.30 s, the tire is stopped and spun in the

opposite (clockwise) direction, also at 3.00 rad/s. Calculate the change in the tire's angular velocity Δω and the tire's averageangular acceleration α. (Indicate the direction with the signs of your answers.)
Physics
1 answer:
emmasim [6.3K]4 years ago
6 0

Answer:

Δω =  -6.00 rad/s

α = -2.61 m/s²

Explanation:

Step 1: Data given

A bicycle tire is spinning counterclockwise at 3.00 rad/s

Δt = 2.30 s

In theopposite (clockwise) direction, also at 3.00 rad/s

Step 2: Calculate the change in the tire's angular velocity Δω

Δω =  ωf -  ωi

ωf = - 3.00 rad/s

 ωi  = 3.00 rad/s

Δω =  ωf -  ωi  = -3.00 - 3.00 = -6.00 rad/s

Step 3: Calculate the tire's average angular acceleration α

α =  Δω /  ΔT

α = -6.00 rad/s /2.30s  

α = -2.61 m/s²

A negative angular acceleration means a decreasing angular velocity

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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
A container has a 5m^3 volume capacity and weights 1500 N when empty and 47,000 N when filled with a liquid. What is the mass de
Helen [10]

Answer:

927.62 kg/m³ and 0.9275.

Explanation:

Density: This can be defined as the ratio of the mass of a body and its volume.

The S.I unit of density is kg/m³.

Mathematically, Density can be expressed as

D = m/v......................... Equation 1

Where D = density of the body, m = mass of the body,v = volume of the body

Also,

m = W/g.................... Equation 2

Where W = weight of the body, g = acceleration due to gravity.

Given: W = 47000 - 1500 = 45500 N, g = 9.81 m/s²

Substitute into equation 2,

m = 45500/9.81

m = 4638.12 kg.

Also given: v = 5 m³

Substitute into equation 1,

D = 4638.12/5

D = 927.62 kg/m³

Hence the mass density of the liquid = 927.62 kg/m³

Specific gravity: This is the ratio of the density of a body to the density pf water.

R.d  = D/D'................................ Equation 3

Where R.d = specific gravity, D = density of the liquid, D' = density of water.

Given: D = 927.62 kg/m³, D' = 1000 kg/m³

Substitute into equation 3

R.d = 927.52/1000

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EastWind [94]

Answer:

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emmasim [6.3K]

Answer:

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Where ε₀ is the dielectric constant and μ₀ the magnetic permittivity.

Let's apply this equation to the present case

 

If we double the electro field

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               u’= 4 u₀

Therefore the energy is multiplied by four

Let's check the answers

a) False

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