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denis-greek [22]

4 years ago

13

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so th

e moment of inertia decreases to 1.80 kg·m2

1 answer:

harkovskaia [24]4 years ago

5 0

Answer:

The angular velocity when the arms are pulled is ω₂= 6.25 rad/s

Explanation:

Assuming that you want to determine the final angular velocity when is the arms are pulled, then it is calculated using the principle of conservation of angular momentum. It states that:

I₁ω₁=I₂ω₂

where I = moment of inertia , ω= angular velocity , 1 and 2 denote the skater with extended hands and pulled respectively.

Thus

I₁ω₁=I₂ω₂

ω₂= I₁ω₁/I₂

replacing values

ω₂= ω₁ *(I₁/I₂) = 5 rad/s *(2.25 kg·m2/1.80 kg·m2) = 6.25 rad/s

ω₂= 6.25 rad/s

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A basebal (radius = .036 m, mas = .145kg) is droped from rest at the top of the Empire State Building (height = 1250ft). Calcula

lora16 [44]

Answer:

a) the initial potential energy = 541.95J

b) the final kinetic energy = 87.991 J

Explanation:

<u>Step 1</u>: Data given

The ball has a mass of 0.145 kg and is at a height of 381m

<u>Step 2</u>: Calculate potential energy

The potential energy = m * g * h

with m = the mass of the ball = 0.145 kg

with g = Gravitational acceleration = 9.81 m/s²

with h = the height of the building = 381m

The potential energy = 0.145 Kg * 9.81 m/s² * 381m = 541.95 J

FD = 1/2ρCDAv²

⇒ with FD = the drag force = the force component in the direction of the flow velocity

⇒ with ρ = density of the fluid (air in our case: ρ≈1.1839 Kg/m3 at 1 atm and 25 °C)

⇒ with v = velocity of the ball

⇒ with A = reference area, which in our case is just the cross sectional area of the ball: A=πr2

⇒ with CD is the drag coefficient - a dimensionless coefficient, that in the case of a sphere, CD=0.47

Following Newton's second law:

ΣFy = may = -mg +Dv²

Here is D=1/2ρCDA ( for convenience) = 0.001172

The terminal speed we can define as the speed of the ball where ay = 0

Therefore: -mg + Dvt² = 0

⇒vt = √(mg/D)

vt = √(0.145 * 9.8 / 0.001172) ≈ 34.837

Uinitial=mgh≈541.951 J (see first question)

Kfinal=1/2*mvt²=(m²g)/2D ≈ 87.991 J

The final kinetics energy is 87.991 J

4 0

4 years ago

Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area

yaroslaw [1]

Answer:

$5.63\cdot 10^{-6} C$

Explanation:

The capacitor of a parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

A is the area of each plate

d is the separation between the plates

$\epsilon_0$ is the vacuum permittivity

The energy stored in a capacitor instead is given by

$U=\frac{1}{2}\frac{Q^2}{C}$

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

$U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}$

And re-arranging it

$Q=\sqrt{\frac{2U\epsilon_0 A}{d}}$

Now if we substitute

$d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J$

We find the charge stored on the capacitor:

$Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C$

7 0

3 years ago

Gold has a specific heat of 129 J/kg°C. How many joules of heat energy are required

ikadub [295]

To raise 1 kg by 1C, it will take 129 joules.

To raise 0.015 kg by (85-22)C, it will take (129*0.015)*22 = 42.57 joules.

6 0

3 years ago

You set a small eraser at a distance of 0.27 m from the center of a turntable you find in the attic and establish that when the

Licemer1 [7]

Answer with Explanation:

We are given that

Distance,r=0.27 m

Tangential speed=v=0.49 m/s

a.Angular speed , $\omega=\frac{v}{r}$

Using the formula

$\omega=\frac{0.49}{0.27}=1.8 rad/s$

$\omega=\frac{2\pi}{T}$

$T=\frac{2\pi}{\omega}$

Time period, $T=\frac{2\pi}{1.8}=3.49 s$

b.Amplitude,A=Distance of small eraser from the center of a turnable =0.27 m

c.Maximum speed, $V_{max}=0.49 m/s$

d.Maximum acceleration= $a=r\omega^2=0.27(1.8)^2=0.87 m/s^2$

5 0

4 years ago

Read 2 more answers

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

B)

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

C)

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

3 0

4 years ago