Does sit-ups reduce belly fat?

Physics

2 answers:

Lemur [1.5K]2 years ago

4 0

Yes it does because it exercises the muscles :))

Naddik [55]2 years ago

3 0

It does. When you are doing sit ups, it makes your core work and if you have a good strong core then you are burning fat. <span />

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A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55

Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

$\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s$

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

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2 years ago

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il63 [147K]

the SL unit of acceleration is the meter per second squared

3 0

2 years ago

Review. A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250-kg block on the left and a 0

photoshop1234 [79]

The answer is 0.245N.

<h3>What is kinetic energy?</h3>

A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.

(b) 0.100

For the block on the left, $f_{k} =u_{k} n= 0.100(2.45N)=0.245N.$

∑ $F_{x}$ = $ma_{x}$

–0.308N+0.245N=(0.250kg)a

a=−0.252 $m/s^{2}$ if the force of static friction is not too large.

For the block on the right, $f_{k}$ = $u_{k} n$ =0.490N. The maximum force of static friction would be larger, so no motion would begin, and the acceleration is zero