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seropon [69]
3 years ago
12

Jessica stretches her arms out 0.60 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins aroun

d on an ice rink at 1.1 m/s.
a. What is the combined angular momentum of the masses?
b. If she pulls her arms into 0.15 m, what is her new linear speed?

I'm really confused ab the explanations behind this. help, please
Physics
1 answer:
Juliette [100K]3 years ago
4 0

Answer:

a.) L = 2.64 kgm^2/s

b.) V = 4.4 m/s

Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.

So,

Radius r = 0.6 m

Mass M = 2 kg

Velocity V = 1.1 m/s

Angular momentum L can be expressed as;

L = MVr

Substitute all the parameters into the formula

L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1

the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1

b. If she pulls her arms into 0.15 m,

New radius = 0.15 m

Using the same formula again

L = 2( MVr)

2.64 = 2( 2 × V × 0.15 )

1.32 = 0.3 V

V = 1.32/0.3

V = 4.4 m/s

Her new linear speed will be 4.4 m/s

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3 years ago
1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com
Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s

d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}

d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}

d' = 594\sqrt{1 - 0.81}

d' = 594 \times 0.4359

d' = 258.92m

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3 0
3 years ago
Block 1, with mass m1 and speed 3.6 m/s, slides along an x axis on a frictionless floor and then undergoes a one-dimensional ela
irina1246 [14]

Answer:

a) The block 1 slides 0.24 m into the rough region.

b) The block 2 slides 2.7 m

Explanation:

Hi there!

First, let´s find the final velocity of each block. With that velocities, we can calculate the kinetic energy of each block. The kinetic energy of the blocks will be equal to the work done by friction to stop them. From the equation of work, we can calculate the distance traveled by the blocks.

Since the collision is elastic, the momentum and kinetic energy of the system composed of the two blocks is constant.

The momentum of the system is calculated as the sum of the momenta of each block:

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

Where:

m1 and m2 = mass of blocks 1 and 2 respectively.

v1 and v2 = velocity of blocks 1 and 2 respectively.

v1´ and v2´ = final velocity of blocks 1 and 2 respectively.

Using the data we have, we can solve the eqaution for v1´:

m1 · 3.6 m/s + 0.40 m1 · 0 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s · m1 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s = v1´ + 0.40 v2´

v1´ = 3.6 m/s - 0.40 v2´

The kinetic energy of the system also remains constant:

1/2 m1 · (v1)² + 1/2 m2 · (v2)² = 1/2 m1 · (v1´)² + 1/2 m2 · (v2´)²

Multiply by 2 both sides of the equation:

m1 · (v1)² + m2 · (v2)² = m1 · (v1´)² + m2 · (v2´)²

Let´s replace with the data:

m1 · (3.6 m/s)² + 0.40 m1 · 0 = m1 · (v1´)² + 0.40 m1 (v2´)²

divide by m1:

(3.6 m/s)² = (v1´)² + 0.40 (v2´)²

Replace v1´ = 3.6 m/s - 0.40 v2´

(3.6 m/s)² = (3.6 m/s - 0.40 v2´)² + 0.40 (v2´)²

Let´s solve for v2´:

(3.6 m/s)² = (3.6 m/s)² - 2.88 v2´ + 0.16 (v2´)² + 0.40 (v2´)²

0 = 0.56 (v2´)² - 2.88 v2´

0 = v2´(0.56 v2´ - 2.88)   v2´ = 0 (the initial velocity)

0 = 0.56 v2´ - 2.88

2.88/0.56 = v2´

v2´ = 5.1 m/s

Now let´s calculate v1´:

v1´ = 3.6 m/s - 0.40 v2´

v1´ = 3.6 m/s - 0.40 (5.1 m/s)

v1´ = 1.56 m/s

Now, let´s calculate the final kinetic energy (KE) of each block:

a) Block 1:

KE = 1/2 · m1 · (1.56 m/s)² = m1 · 1.2 m²/s²

The work done by friction is calculated as follows:

W = Fr · s

Where:

Fr = friction force.

s = traveled distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction.

And the normal force is calculated in this case as:

N = m1 · g

Where g is the acceleration due to gravity.

Then, the work done by friction will be:

W = m1 · g · μ · s

The kinetic energy of an object is the negative work that must be done on that object to bring it to stop. Then:

m1 · 1.2 m²/s² = m1 · g · μ · s

Solving for s:

s = m1 · 1.2 m²/s²  / m1 · g · μ

s = 1.2 m²/s²/ 9.8 m/s² · 0.50

s = 0.24 m

The block 1 slides 0.24 m into the rough region.

b) For block 2 the kinetic energy will be the following:

KE = 1/2 · 0.4 · m1 · (5.1 m/s)² = m1 · 5.2 m²/s²

The friction force will be:

Fr = 0.4 m1 · g · μ

And the work done will be:

W = 0.4 m1 · g · μ · s

Since W = ΔKE,

Then:

m1 · 5.2 m²/s² = 0.4 m1 · g · μ · s

Solving for s:

5.2 m²/s²/(0.4 · g · μ) = s

s =  5.2 m²/s²/(0.4 · 9.8 m/s² · 0.50)

s = 2.7 m

The block 2 slides 2.7 m

3 0
3 years ago
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