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Pavlova-9 [17]
3 years ago
15

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end

is allowed to fall. What is the angular acceleration as it is released
Physics
1 answer:
Natalka [10]3 years ago
5 0

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod  and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

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Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
sveta [45]

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two  or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with m_{Pa} and its initial velocity with u_{Pa}

Let represent the mass of Ricardo with m_{Ri} and its initial velocity with u_{Ri}

At rest ;

their velocities will be zero, i.e

u_{Pa} = u_{Ri} = 0

The initial momentum for this process can be represented as :

m_{Pa}u_{Pa} +  m_{Ri}u_{Ri} = 0

after push off from each other then their final velocity will be v_{Pa} and v_{Ri}

The we can say their final momentum is:

m_{Pa}v_{Pa} +   m_{Ri}v_{Ri} = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

m_{Pa}u_{Pa} +  m_{Ri}u_{Ri} =  m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}

Since the initial velocities are stating at rest then ; u = 0

m_{Pa}(0) + m_{Pa}(0) = m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}

m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}  = 0

m_{Pa}v_{Pa} = - m_{Ri}v_{Ri}

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

 B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So m_{Ri} > m_{Pa} ;

Then \mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

m_{Pa}v_{Pa} =  m_{Ri}v_{Ri}

The ratio is

\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1

v_{Pa} >v_{Ri}

Therefore, Paula will have a greater speed than Ricardo after the push-off.

6 0
3 years ago
Line d represents water. if the atmospheric pressure in a flask is lowered to 70 kpa, water would boil at what temperature?
sp2606 [1]
The diagram is in the picture attached.
Options are:
A) 32 °C
B) 70 °C
C) 92 °C
D) 100 °C

In order to find the value required, you need to look at the diagram and follow these steps:
1) search for the value of 70 kPa on the y-axis;
2) move on a horizontal line towards the right until you reach the line D;
3) move on a vertical line down, towards the x-axis;
4) read at what value of °C you are at.

Doing so, you can see that you are at a value a little bit above 90 °C (see picture).

Hence, the correct answer is C) 92°C.

5 0
3 years ago
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