All categories

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<span> have less friction so there is a lesser chance of </span>slipping<span>. 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</span>

You might be interested in

P2O5 is a covalent compound used to purify sugar. What is the name of this compound?

AfilCa [17]

Answer:

B) Diphosphorus pentoxide

Explanation:

8 0

3 years ago

Read 2 more answers

Would it be better to collide head-on with an identical car traveling at the same speed or collide with a stationary brick wall?

attashe74 [19]

It makes no difference. The momentum of either car goes to zero in both cases.

6 0

3 years ago

A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the

ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

$\frac {Vf^{2}-Vo^2}{2.a} =X$

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

$Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s$

the speed with the ball hits the ground is 32.46m/s

8 0

3 years ago

Read 2 more answers

You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The

LuckyWell [14K]

Answer: $4.65\ m/s$

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

$v^2-u^2=2as$

$\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s$

So, the velocity of putty just before hitting is $4.65\ m/s$

5 0

3 years ago

Two ice skaters stand together as illustrated in the attached figure. They "push off" and travel directly away from each other,

Tema [17]

By definition, the momentum is given by:
p = m * v
Where,
m = mass
v = speed.
On the other hand,
F = m * a
Where,
m = mass
a = acceleration:
For the boy we have:
p1 = m * v
p1 = (F / a) * v
p1 = ((710) / (9.81)) * (0.50)
p1 = 36.19 Kg * (m / s)
For the girl we have:
p2 = m * v
p2 = (F / a) * v
p2 = ((480) / (9.81)) * (v)
p2 = 48.93 * v Kg * (m / s)
Then, we have:
p1 + p2 = 0
36.19 + 48.93 * v = 0
Clearing v:
v = - (36.19) / (48.93)
v = -0.74 m / s (negative because the velocity is in the opposite direction of the boy's)
Answer:
the girl's velocity in m / s after they push off is -0.74 m / s

6 0

3 years ago