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hichkok12 [17]
3 years ago
15

A bucket of water can be whirled in a vertical circle without the water spilling out, even at the top of the circle when the buc

ket is upside down. explain.
Physics
2 answers:
Andrei [34K]3 years ago
3 0
<span>This is actually all because of centripetal acceleration and inertia. You see that when the bucket spins, it actually tries to go straight, however the rope attached restricts this and makes it go in a circle. This makes the water to be pinned against the bottom of the bucket. </span>
Vedmedyk [2.9K]3 years ago
3 0

Answer:

centripetal force

Explanation:

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madam [21]

Answer:

μ=0.151

Explanation:

Given that

m= 3.5 Kg

d= 0.96 m

F= 22 N

v= 1.36 m/s

Lets take coefficient of kinetic friction = μ

Friction force Fr=μ m g

Lets take acceleration of block is a m/s²

F- Fr = m a

22 -  μ x 3.5 x 10 = 3.5 a         ( take g =10 m/s²)

a= 6.28 - 35μ  m/s²

The final speed of the block is v

v= 1.36 m/s

We know that

v²= u²+ 2 a d

u= 0 m/s given that

1.36² = 2 x a x 0.96

a= 0.963 m/s²

a= 6.28 - 35μ  m/s²

6.28 - 35μ = 0.963

μ=0.151

3 0
3 years ago
John has decided he needs to work harder on his Social Studies project. Then, his teacher says if he gets a good grade in Social
bezimeni [28]

Answer: Extrinsic to intrinsic

Explanation:

8 0
3 years ago
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A force pointing in the xx-direction is given by F=ax3/2F=ax3/2, where aa is a constant. The force does 2.01 kJkJ of work on an
damaskus [11]

Answer:

Explanation:

Given that

F=ax^3/2. a is a constant

The force does a work of

W=2.01KJ from x=0 to x=15.2m

We need to find a

Work is give as,

W=∫F.ds

But this is in x direction only then,

W=∫Fdx. from x=0 to x=15.2m

W=∫ax^3/2dx from x=0 to x=15.2m

W=ax^(3/2+1)/(3/2+1).

W=ax^(5/2)/5/2

W=ax^(2/5)/2.5 from x=0 to x=15.2m

Cross multiply

2.5W=ax^2.5. from x=0 to x=15.2m

2.5W= a (15.2^2.5-0)

W=2.01KJ=2010J

2.5×2010=a×900.76

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a=5.56

7 0
3 years ago
Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri
Elanso [62]

First we can say that since there is no external force on this system so momentum is always conserved.

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}

0.48= 0.6*v_{1f} + 1.2*v_{2f}

0.8  = v_{1f} + 2v_{2f}

now by the condition of elastic collision

v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6

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also from above equation we have

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So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

8 0
3 years ago
Is the answer correct or wrong ​
Zielflug [23.3K]

Answer:

wrong answer because m=100

7 0
3 years ago
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