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3 years ago

Explain why it is easy to slip on a floor that is wet please, I need this answered now if anyone can.

2 answers:

5 0

It is easy to slip on a floor that is wet because there is less friction between your shoe/foot and the floor. Friction is basically when two parallel objects come in contact and form resistance. An effect of friction is the slowing down of velocity in the moving object aka your foot. This is why a rocky surface is less slippery than a marbled surface. Likewise with a west and dry surface. Hope this helps!

VladimirAG [237]3 years ago

4 0

Wet floors have less friction so there is a lesser chance of slipping. This is because the coefficient of friction μ and the therefore resultant force of limiting friction Fs is significantly reduced when a lubricant is introduced between the the two kinetic or static surfaces

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A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instan

gizmo_the_mogwai [7]

Answer:

The velocity just before hitting the ground is $v_f = 30 m/s$

Explanation:

From the question we are told that

The initial speed is $u = 10 m/s$

The final speed is $v = 30 \ m/s$

From the equations of motion we have that

$v^2 =u^2 + 2as$

Where s is the distance travelled which is the height of the cliff

So making it the subject of the the formula we have that

$s = \frac{v^2 - u^2 }{2a}$

Where a is the acceleration due to gravity with a value $a = 9.8m/s^2$

$s = \frac{30^2 - 10^2 }{2 * 9.8 }$

$s = 40.8 \ m$

Now we are told that was through horizontally with a speed of

$v_x =10 m/s$

Which implies that this would be its velocity horizontally through out the motion

Now it final velocity vertically can be mathematically evaluated as

$v_y = \sqrt{2as}$

Substituting values

$v_y = \sqrt{(2 * 9.8 * 40.8)}$

$v_y = 28.3 \ m/s$

The resultant final velocity is mathematically evaluated as

$v_f = \sqrt{v_x^2 + v_y^2}$

Substituting values

$v_f = \sqrt{10^2 + 28.3^2}$

$v_f = 30 m/s$

5 0

3 years ago

Area is found by multiplying the length of a surface times the width. If a floor measures 5.28 m2, how many square centimeters d

Scilla [17]

62,500 cm^2.........

8 0

3 years ago

Um relógio de ponteiros funciona durante um mês. Qual é, em N.C, o número de voltas, aproximadamente, que o ponteiro dos minutos

Margaret [11]

I can’t understand I’m sorry but if you need help with math I can help you

7 0

3 years ago

The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS

deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

$v=\frac{2\pi r}{T}$ (1)

where

$2\pi r$ is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: $v=7,750 m/s$ . The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface: