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3 years ago

Can someone write and short summary on how to hula hoop

2 answers:

5 0

You pick up the hula hoop and stand inside of it then pick it up will you’re inside it and hold to your waist and spin it then turn your hips

MissTica3 years ago

5 0

Answer:

First you take a hula hoop and hold it around your waist then you start moving your hips in a circle and let the hula hoop go off your hands and you keep moving your hips. You keep moving your hips until you want to stop but it might fall. Just keep trying and you should get the hang of it in a few minutes.

Explanation:

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A 70-kg boy is sitting 3 m from the ground in a tree. What is his gravitational potential energy

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m x h x 9.8 m/s squared

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3 years ago

Which sentence states Newton’s third law?

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Answer:

A. If two objects collide, each object exerts a force in the same direction as the other.

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A book sitting on a table is moved horizontally. Describe the

Akimi4 [234]

Frictional force and Applied force has same “magnitude” and “opposite” direction.

Option: B

<u>Explanation</u>:

When a book is moved horizontally by applying “force” on the book, the frictional force is opposed to the book by the table. Here, this “frictional force” is opposing the book has the same force what we applied on the book but this frictional force and the applied force are opposite in direction. Always the “frictional force” is opposite to the “applied force” which stops the object to move. For example, if a force applied leftward to the object the frictional force is acted on the right side of the object.

When two objects are in contact they experience a "frictional force". This "frictional force" acts opposite to the force applied on to move the object.

Formula for "frictional force" is $\mu\times N$

Where, $\mu$ is coefficient of friction and N is normal force.

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3 years ago

Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a

Naya [18.7K]

a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

$p_i = m u = (8 kg)(2 m/s)=16 kg m/s$

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

$p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s$

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

$p_f = p_{fB}+p_{fS}$

where $p_{fS}$ is the momentum of the spring. For the conservation of momentum,

$p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s$

b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

$\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s$

c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

$\Delta p=F \Delta t$

Since we know $\Delta t=0.5 s$ , we can find the magnitude of the force:

$F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N$

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.

d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

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3 years ago