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ExtremeBDS [4]
4 years ago
5

two children with same mass in a merry-go-round are at different distance from the center, child 'A' closer than 'b'. which one

takes longer to complete one revolution?
Physics
2 answers:
Murljashka [212]4 years ago
4 0

Answer:

<em>Both take the same time to complete one revolution</em>

Explanation:

<u>Circular Motion</u>

Imagine you and your friend are playing in the same merry-go-round. If he is sitting closer to the center that you, it won't make him move with respect to yourself. You are always in the same relative position from each other. If you complete a revolution, say, in 10 seconds, then your friend will also complete the revolution in 10 seconds.

Since both children are in different positions respect to the center and they complete the revolutions simultaneously, then they necessarily have different velocities. The closer to the center the child is, the less velocity he will have.

We are talking about the tangent velocity or vt. The other velocity is the angular velocity and that is the same for both children.

SSSSS [86.1K]4 years ago
4 0

Answer:

Explanation:

Let the distance of child A is r and the distance of child B is R.

Let the mass of each student is m.

According to the conservation of angular momentum

I_{A}\omega _{A}=I_{B}\omega _{B}

Let the time taken by A is TA and the time taken by B is TB.

mr^{2}\times \frac{2\pi }{T_{A}}=mR^{2}\times \frac{2\pi }{T_{B}}

\frac{T_{B}}{T_{A}}}=\left ( \frac{R^{2}}{r^{2}} \right )

As R > r

So, TB > TA

Thus, the time taken by child B is more than the child A.

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What is the function of the organelle labeled E in the diagram?
nlexa [21]

Answer: D. Storage spaces in the cell.

Explanation: The organelle labeled E is called a vacuole and it’s used for storage in both plant and animal cells.

7 0
4 years ago
What does area under a velocity time graph represent
Feliz [49]
Velocity hope that helps
7 0
3 years ago
A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui
raketka [301]

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

8 0
4 years ago
A bullet glider and a target glider both have a mass of 0.200 kg. The bullet glider is moving 0.450 m/s
Romashka [77]

Answer:

the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Explanation:

When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.

The whole process must be analyzed using conservation of the moment.

             p₀ = m v₀

celestines que clash case

             p_f = (m + M) v

             po = pf

             m v₀ = (n + M) v

             v = \frac{m}{m+M}

calculemos

            v= \frac{0.200}{0.200+M} 0.450

            v= 0.09 m/s

elastic shock case

           p₀ = m v₀

           p_f = m v₁ +M v₂

           p₀ = p_f

           m v₀ = m v₁ + m v₂

6 0
3 years ago
a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?​
lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let f be frictional force acting in the backward direction as shown. Let the coefficient of friction be \mu. Let N be the normal reaction force acting on the bag.

Given:

Mass of the bag is, m=8.10\textrm{ kg}

Force acting at \theta = 38° is F= 29.5\textrm{ N}

Acceleration due to gravity is, g=9.8\textrm{ }m/s^{2}

The force F can be resolved into its components as F_{x}=F \cos \theta and F_{y}=F \sin \theta

Therefore,

F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

f=F_{x}\\f=23.25\textrm{ N}

Now, frictional force is given as:

f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38

Therefore, the coefficient of friction is 0.38.

8 0
4 years ago
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