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34kurt
3 years ago
12

1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour

se. If the hoop has a radius of 75.0 cm, what is the moment of inertia of Alex's rolling hoop?
a) 1,970 kg ⋅ m2
b) 26.3 kg ⋅ m2
c) 0.263 kg ⋅ m2
d) 0.197 kg ⋅ m2

2. Jessica stretches her arms out 0.8 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins around on an ice rink at 1.2 m/s.
a. What is the combined angular momentum of the masses?
b. If she pulls her arms in to 0.12 m, what is her new linear speed?
Physics
1 answer:
grigory [225]3 years ago
6 0

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 kg \cdot cm^2

Explanation:

<u>Given</u>:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

<u>To Find:</u>

The moment of inertia of Alex's rolling hoop = ?

<u>Solution</u><u>:</u>

The moment of inertia  = mr^2

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia  = (0.350)(0.75)^2

=> moment of inertia  = (0.350)(0.5625)

=> moment of inertia  = (0.197)

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 kg m^2 s^{-1}

If she pulls her arms in to 0.12 m, her new linear speed  is  18.33 m/s^2

Explanation:

Given:

Mass  = 2.0 kg

Radius = 0.8 m

Velocity =  1.2 m/s

a.The combined angular momentum of the masses:

L = r \cdot m \cdot v_1

Substituting the values,

L = 0.8 \cdot 2.0 \cdot 1.1

L= 1.76 kg m^2 s^{-1}

b. If she pulls her arms in to 0.12 m, what is her new linear speed

0.12 \cdot 0.8 \cdot v_2 = 1.76

0.096 cdot v_2 = 1.76

v_2 = \frac{1.76}{0.096}

v_2 = 18.33 m/s^2

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1 point
s2008m [1.1K]

Answer:

The person has no displacement

Explanation:

The given parameters are

The location of the person = The equator

The distance covered in one revolution = Total distance around the body

The total distance around the Earth = The circumference of the Earth = 40.075 kilometres

The total distance moved by the person standing at the equator during the Earths complete revolution = 40,075 kilometres

The initial location of the person in relation to a fixed point in space outside Earth at the start of the revolution = x km

The final location of the person in relation to the fixed point in space outside Earth at the completion of the revolution = x km

The displacement = Change in position = Final location - Initial location  

∴ The displacement = x km - x km = 0 km.

5 0
4 years ago
A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi
bogdanovich [222]

Answer:

1) W_{object} = 400 N

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

\sum F_{Tot} = 0

In this case we just have y-direction forces.

\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0

Now, let's solve the equation for W(object).

W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N

2 ) To find the position of the heaviest weight we need to use the torque definition.

\sum \tau = 0

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

L = 1.50 m  ; total length of the bar.

D_{AP} = x  ; distance between Tension A and P point.

D_{BP} = L-x ; distance between Tension B and P point.

D_{W_{bar}P} = \frac{L}{2}-x ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0

Finally we just need to solve it for x.

x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}

x = 0.28 m

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0
4 years ago
HELP!!!!
BigorU [14]
Time = 13.5 / 2.5 = 5.4 seconds
4 0
3 years ago
Read 2 more answers
How far can a person run in 15 minutes if he or she runs at an average speed of 16 km/hr
Studentka2010 [4]
4 km/hr. There are four quarters in an hour, so just divide 16 by 4 and you get 4. It is directly proportional.
6 0
3 years ago
the young’s modulus of aluminum is 69gpa, of nylon is 3gpa, of tungsten is 400gpa, and of copper is 117gpa. if equal-size sample
olga_2 [115]

Answer:

Least to most elongated: tungsten, copper, aluminum, nylon.

Explanation:

Materials with high Young's modulus are difficult to stretch. σ = Yε and ε = ΔL/L so an object with a high Young's modulus (Y) subject to a certain tensile stress (σ) will have a smaller strain than an object with a smaller Young's 's modulus subject to the same tensile stress. If strain (ε) is smaller, then ΔL will also be smaller.

3 0
2 years ago
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