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maria [59]
3 years ago
15

Exactly how fast would the average human have to run in order to actually ran across water without breaking surface tension? Thi

s is just a little thing I'm curious about and would like to see if anyone knows the answer.
Physics
2 answers:
Mila [183]3 years ago
7 0

Answer:

It wouldn't work no matter how fat a person runs

Explanation:

The reason for this is because of the weight of a human and that a human can't possibly get to that kind of speed.

lina2011 [118]3 years ago
7 0

Answer:

It's something weird like  67 mph

Not sure if that's exactly right, but i hope it helps

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Calculate the potential V(r) for r>rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.)
Marat540 [252]

Answer:

The potential for r > rb is equal to zero.

Explanation:

For r > rb, the potential is:

V=\frac{Kq}{r}

Then, the net potential is:

V_{(r)} =\frac{K(+\epsilon )}{r} +\frac{K(-\epsilon )}{r}

K=\frac{1}{4\pi \epsilon _{o}  }

V_{(r)} =\frac{K(+\epsilon )}{r} -\frac{K(\epsilon )}{r}\\V_{(r)}=0

8 0
4 years ago
Why would a person who weighs 100 lbs. on Earth weigh significantly more on Ju-piter?
Nitella [24]

Answer:

Is a

Explanation:

8 0
3 years ago
Electrons move towards the ________ end.<br> O negative<br> O positive<br> O front<br> O back
LUCKY_DIMON [66]

Answer:

they move towards the positive side... that's option 2

8 0
3 years ago
Read 2 more answers
A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi
NeX [460]

Answer:

h = 2.087 m

Explanation:

Given

m₁ = 3 kg

v₁ = 20 m/s

m₂ = 2 kg

v₂ = - 14 m/s

In a completely inelastic collision the colliding objects stick together after the collision and move together as a single object.

In the given problem, lets assume that the balls of putty are initially moving along the  y  axis, upward direction being the positive  y  direction. And the collision occurs at the origin of the coordinate system.

We can apply the equation

vs = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)  

⇒   vs = (3 kg*20 m/s + 2 kg*(- 14 m/s)) / (3 kg + 2 kg)  

⇒   vs = 6.4 m/s (↑)

To calculate the maximum height  h  attained by the combined system of two balls of putty after the the collision, we use the expression for linear motion under gravity:

vf² = vi² - 2*g*h

where

vf = 0 m/s  

g = 9.81 m/s²

vi = vs = 6.4 m/s

finally we get h:

h = vi² / (2*g)

⇒   h = (6.4 m/s)² / (2*9.81 m/s²) = 2.087 m

6 0
3 years ago
A falling ball of mass 0.5 kg experiences a downward force due to gravity of mg (where g = 9.8 m/s2) and an upward force of air
schepotkina [342]

Applying Newton's Second Law of Motion, the acceleration of the ball is 16.8 m/s^2

<u>Given the following data:</u>

  • Mass = 0.5 kg
  • Acceleration due to gravity = 9.8 m/s^2
  • Upward force = 3.5 N.

To find ball's acceleration, we would apply Newton's Second Law of Motion:

First of all, we would determine the net force acting on the ball.

Net \; force = Upward\;force + Downward\;force

Downward\;force = 0.5 × 9.8

Downward force =  4.9 N

Net \; force = 3.5 + 4.9

Net force = 8.4 N

Mathematically, Newton's Second Law of Motion is given by this formula;

Acceleration = \frac{Net\;force}{Mass}\\\\Acceleration = \frac{8.4}{0.5}

<em>Acceleration = 16.8 </em>m/s^2<em />

Therefore, the acceleration of the ball is 16.8 m/s^2

Read more here: brainly.com/question/24029674

6 0
3 years ago
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