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GenaCL600 [577]
3 years ago
9

A good alternative to sit ups is A. Crunches B. Leg Raises C. Side Twists

Physics
1 answer:
astraxan [27]3 years ago
6 0
It would probably be B
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A boy takes hold of a rope to pull a wagon (m = 50 kg) on a surface with a static coefficient of friction μS = 0.25. Calculate t
vivado [14]

Answer:

<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction  opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move  can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

          μ is the coefficient of friction ( μs, in this case, static friction);

          m  is mass of the object and;

          g is the acceleration due to gravity( a constant equal to 9.81 m/s^{2})

from the equation we are provide with;

       μs  = 0.25

       m = 50 kg

       g =  9.81 m/s^{2}

      F =?

Using equation 1

F = 0.25 x 50 kg x  9.81 m/s^{2}

F = 122.63 N  

<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>

6 0
4 years ago
Read 2 more answers
10. A 12 kg load hangs from one end of a rope that passes over a small frictionless pulley. A 15 kg counterweight is suspended f
nekit [7.7K]

Answer:

6

Explanation:

6 0
3 years ago
Im stuck on number 4, am I just supposed to talk about theory because there is no given info.
Strike441 [17]

Explanation:

Assuming no air resistance, the horizontal acceleration at all points is 0, and the vertical acceleration at all points is -g (if near the surface of the Earth, -9.8 m/s²).

6 0
4 years ago
What is the purpose of mucilage in the spaces around the cells of hornworts? Why is this important?
Bumek [7]

New Zealand hornworts and general information on them. ... in the top of the plant and remain attached to it and continue to grow throughout its life. ... are that the spaces between the cells are filled with mucilage rather then air chambers

7 0
3 years ago
A typical running track is an oval with 74-m-diameter half circles at each end. A runner going once around the track covers a di
nirvana33 [79]

Answer:

The acceleration towards the center will be 0.43\ m/s^2

Explanation:

Given the running track is an oval shape, and the diameter of each half-circle is 74 meters.

Also, the runner took 1 minute and 40 seconds to complete 400 m one round of the track.

We need to find the acceleration towards the center.

First, we will find the speed.

v=\frac{d}{t}

Where v is the speed.

d is the distance covered by the rider that is 400 meters.

t is the time taken by the rider to complete the lap which is 1 minute and 40 seconds.  (60\ s +40\ s) = 100 seconds.

So,

v=\frac{400}{100}=4\ m/s

And

a_c=\frac{v^2}{r}

Where a_c is the acceleration towards the center.

r is the radius which will be the half of the diameter 74 meters.

Hence, the radius will be 37 meters.

a_c=\frac{4^2}{37} \\a_c=\frac{16}{37}=0.43\ m/s^2

So, the centripetal acceleration of the rider will be 0.43\ m/s^2

4 0
3 years ago
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