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WITCHER [35]
3 years ago
12

As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94

rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Physics
1 answer:
lapo4ka [179]3 years ago
7 0

Answer:

I₂/I₁ = 0.53

Explanation:

During the motion the angular momentum of the skater remains conserved. Therefore:

Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms

L₁ = L₂

but, the formula for angular momentum is:

L = Iω

Therefore,

I₁ω₁ = I₂ω₂

I₂/I₁ = ω₁/ω₂

where,

I₁ = Initial Moment of Inertia

I₂ = Final Moment of Inertia

ω₁ = Initial Angular Velocity = 3.14 rad/s

ω₂ = Final Angular velocity = 5.94 rad/s

Therefore,

I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)

<u>I₂/I₁ = 0.53</u>

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Each blade of a fan has a radius of 11 inches. If the fan’s rate of turn is 1440o /sec, find the following. (a) The angular spee
notka56 [123]

Answer:

a) 24.43 radians per second

b) 268.73 inches per second

Explanation:

a) The angular speed of the fan on Celsius degrees/second is 1400, so we should convert that value to radians using the fact that 2π rad = 360 °C:

\omega = 1400\frac{C}{s}=1400\frac{C}{s}*\frac{2\pi\,rad}{360\,C}

\omega = 1400\frac{C}{s}=24.43\frac{rad}{s}

b) Linear speed on a point of the blade is related with angular speed of the fan by the equation

v=\omega r

with v linear speed, ω angular speed and r the radius of the blades. So:

v=(24.43\frac{rad}{s})(11 in)

Radians isn't really a unity; it is dimensionless so we can put it or not. So:

v=268.73\frac{in}{s}

3 0
3 years ago
What is the momentum of a man of mass 10kg when he walks with a uniform velocity of 2m/s.
Stolb23 [73]

Answer:

Momentum(p) = 20kgm/s

Explanation:

Given

Mass = 10kg

Velocity = 2m/s

Required

Calculate the momentum of the man

Momentum is calculated as thus

Momentum(p) = Mass(m) * Velocity(v)

or

p = mv

So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;

The formula becomes

Momentum(p) = 10kg * 2m/s

Momentum(p) = 10 * 2 * kg * m/s

Momentum(p) = 20kgm/s

Hence, the momentum of the man is 20kgm/s

5 0
3 years ago
The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b
ZanzabumX [31]

Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})

where;

Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²

k is the dielectric constant = 2.3

b is the outer radius of the conductor = 3 mm

a is the inner radius of the conductor = 0.8 mm

λ is the linear charge density = 18 x 10⁶ V/m

Substitute in these values in the above equation;

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) =  \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V

8 0
3 years ago
In an experiment refractive index of glass was observed to be 1.45,1.56,1.54,1.44,1.54and1.53. Calculate mean value of refractiv
Sever21 [200]
The mean may be calculated by summing the values of the refractive index and dividing the sum by the number of experiments. This is:
Mean = (1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53)/6
Mean = 1.51

The mean absolute error is the sum of the absolute values of errors divided by the number of trials:
MAE = (|1.45-1.51|+|1.56-1.51|+|1.54-1.51|+|1.44-1.51|+|1.54-1.51|+|1.53-1.51|)/6
MAE = 0.043

The fractional error is the MAE divided by the actual value:
Fractional error = 0.043 / 1.51
Fractional error = 43/1510

The percentage error is the fractional error multiplied by 100:
Percentage error = 2.85%
7 0
3 years ago
Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second. It hits stationary ball B and they undergo
blagie [28]
<span>If Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second, and It hits stationary ball B and they undergo elastic collision, thus the two balls have different masses, then the following statement which is true is the statement that stated that there was no y-momentum initially.</span>
7 0
3 years ago
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