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Aleonysh [2.5K]
2 years ago
14

13. How much work does a 35 N force do to lift a potted plant from the floor to a shelf 6 m high?

Physics
1 answer:
Nataly_w [17]2 years ago
8 0

Answer:

W= 210 N

Explanation:

Just use work = Fparallel*d

W= 35*6

W= 210 N

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Function Of The glottis​
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Their function is to produce sound by allowing the free edges of the folds to vibrate against one another and also to act as the laryngeal sphincter when they are closed.

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If the emf produced in a wire is 0.88 volts and the wire moves perpendicular to a magnetic field of strength 0.075 newtons/amper
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Emf = d (phi-B) / dt 
<span>B dA/dt, where dA/dt is the area swept out by the wire per unit time. </span>
<span>0.88 V = (0.075 N/(A m)) (L)(4.20 m/s), so </span>
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A dog walks 12 meters to the east and then 16 meters back to the west for this motion what is the distance moved What is the mag
FinnZ [79.3K]

The distance is 28 meters, and the displacement is -4.

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would really appreciate a brainliest! Hope this helped!

6 0
3 years ago
A(n) ________ is a specialized bursa that wraps around a tendon to reduce friction.
Kisachek [45]

Tendon Sheath - is a specialized bursa that wraps around a tendon to reduce friction.

<h3>What is Tendon Sheath ?</h3>

Tendon Sheath is a thin layer of tissue, surrounds each tendon in our body. The tendon sheath can also be called synovial lining or fibrous sheath. Tendon sheaths help to protect tendons from abrasive damage as they move.

Connection between Bursa and Tendon Sheath : Bursae are small fluid-filled sacs that can lie under a tendon, cushioning the tendon and protecting it from the injury. Bursae also provides an extra cushioning to adjacent structures that otherwise might rub against each other, which will cause wear and tear ( example, between a bone and a ligament ) .

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To know more about Tendon Sheath please click here : brainly.com/question/17087116

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8 0
11 months ago
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp
Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J

So, the new work is more than 130 J.

6 0
3 years ago
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