I need a equation for a problem in using a pulling force a mass a coefficient of static friction a coefficient of kinetic friction and solving for the frictional force

Answers

Answer 1
Answer: The minimum pulling force required to start it moving is
equal to the static frictional force.

F = (mass) x (gravity) x (coefficient of static friction).


Once it's moving, the force required to keep it moving at a steady speed,
without speeding up or slowing down, is the force of kinetic friction.

F = (mass) x (gravity) x (coefficient of kinetic friction).


In both equations, [ (mass) x (gravity)] is just the object's 'weight'.




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The pain in the leg that is referred to as "shin splints" is often caused by microtears in the periosteum and perforating fibers. These tears lead to inflammation of the periosteum. Considering the type of tissue found in the periosteum, which cells do you think would be most involved in the repair process?

Answers

Answer:

fibroblast cells

Explanation:

Periosteum is thin connective tissue layer covering the bone. It contains fibroblast cells outside and progenitor cells inside. So, the cells most involved in the repair is fibroblast cells. It is typical type of biological cell that synthesizes matrix and collagen, produces the stroma for animal tissues and plays vital role in healing wounds

what kind of wave would be observed in outer space between planets where there is very little matter?

Answers

Shock waves can be observed in outer space between planets where there is very little matter. Although little, plasma does exist in outer space which allows the waves to travel.

Answer:

Either A or C

Explanation:

there are 2 quizlets with 2 different answers, i don't know, man.

A dolphin leaps out of the water at an angle of 36.6° above the horizontal. The horizontal component of the dolphin's velocity is 7.87 m/s. Calculate the magnitude of the vertical component of the velocity.

Answers

Answer:

Explanation:

Given

Dolphin leaps out an angle \theta =36.6^(\circ)

Horizontal component of dolphin velocity u_x=7.87\ m/s

Suppose u is the launch velocity of dolphin

therefore u\cos \theta =u_x---1

and vertical velocity u_y=u\sin \theta ----2

divide 1 and 2 we get

\tan \theta =(u_y)/(u_x)

u_y=u_x\tan \theta

u_y=7.87\cdot \tan (36.6)

u_y=5.84\ m/s

A total solar eclipse is visible froma. all over Earth.
b. only within the moon’s umbra.
c. only within the moon’s penumbra.
d. only the dark side of Earth.

Answers

Answer:

b. only within the moon’s umbra.

Explanation:

Solar eclipse occurs when Moon comes between the Sun and the Earth in straight line and blocks the sunlight. A shadow of moon castes on the surface of the Earth. The region passing through the moon's umbra would view a total solar eclipse. The moon completely hides the surface of the sun. It would be visible to the people residing in the bright side of the Earth within the Umbra of the moon.

Only within the umbra of the moon's shadow.

A bookshelf is 2m long, with supports at its end (p and q). A book weighing 10N is placed in the middle of the shelf. what are the upward forces acting on p and q. if the book is moved 50 cm from q, what are the forces at p and q now?

Answers

When the book is placed in the middle, the forces acting on p and q is 5N. When the book is moved 50 cm from q, the forces at p and q can be solved by doing a moment balance
With p as the pivot
Fq (2 m) = 10 N (0.5 m)
Fq = 2.5 N
and
Fp = 10 N - 2.5 N = 7.5 N 

Answer:

Fq = 2.5N

Fp = 7.5N

Explanation:

Hello! Let's solve this!

When the book is in the middle, the force in p (Fp) and the force in q (Fq) are equal, and half of the total force

Fp = Fq = 5N

When the book moves 0.5m from q we have a balance

Fq * 2m = 10N * 0.5m

We cleared Fq

Fq = (10N * 0.5m) / 2m

Fq = 2.5N

Fp = 10N-Fq = 10N-2.5N = 7.5N

Conclusion

Fq = 2.5N

Fp = 7.5N

A radioactive substance is decaying exponentially. if there are initially 15 grams of the substance and after 120 minutes there are 4 grams, what is the half-life in minutes of the substance?

Answers