What kind of friction exists between solid objects moving in water?

Physics

1 answer:

eduard3 years ago

6 0

Fluid Friction exists when it is acted upon an object when in fluid.

You might be interested in

Larry and Balky set up an experiment to analyze the motion of a marble as it rolled down an incline. They started the marble at

erik [133]

Answer:

Option A is the correct answer.

Explanation:

The instantaneous acceleration = Change in velocity in velocity/Time taken

The slope of the graph should give instantaneous acceleration.

Slope of a graph = Change in value of Y -axis / Change in values of X -axis

Comparing both the equations

Change in value of Y -axis = Change in velocity in velocity

Change in values of X -axis = Time taken

So velocity values should be on the Y axis and Time values should be on the X axis.

Option A is the correct answer.

7 0

3 years ago

Read 2 more answers

You are in a hot air balloon (yes, another balloon problem!) rising from the ground at a constant velocity of 2.00 m/s upward. t

Daniel [21]

If all you need is the initial speed of the cork, you can solve this using only two of your given:
2.00 m.s upward and 6.60 m.s horizontally.

If you take in consideration the movement of the cork, you know that it was both going up and forward at the same time, this means that it was moving at a diagonal direction. Now you can solve this by using the Pythagorean theorem where:

$c = \sqrt{ a^{2} + b^{2} }$

Why? Because the vertical and the horizontal motion creates a movement that is diagonal, which when put in a free-body diagram, creates a right triangle.

Going back to your problem, when applying this, the diagonal of a right triangle is the hypotenuse, so this is what you are looking for. The horizontal and vertical motion will represent the other 2 sides of the triangle.

Now let's put that into your formula:

$c = \sqrt{ a^{2} + b^{2} }$

$Vi = \sqrt{ Vx^{2} + Vy^{2} }$

Where: Vx is your horizontal velocity
Vy is your vertical velocity
Vi is your initial velocity

Now let's put in your given:

$Vi = \sqrt{ Vx^{2} + Vy^{2} }$
$Vi = \sqrt{ 6.60^2} + 2.00^{2} }$
$Vi = \sqrt{ 43.56 + 4.00 }$
$Vi = \sqrt{ 47.56 }$
$Vi = 6.8964 m/s$

So your initial velocity is 6.8964 m/s or 6.90 m/s

5 0

4 years ago

At the presentation ceremony, a championship bowler is presented a 1.60-kg trophy which he holds at arm's length, a distance of

nikdorinn [45]

Answer:

a)10.28 Nm

b)9.93 Nm

Explanation:

Let g = 9.81m/s2. First we can calculate the weight of the trophy

W = mg = 1.6 * 9.81 = 15.696 N

(a) The torque is product of force and its moment arm

T = WL = 15.696 * 0.655 = 10.28 Nm

(b) Suppose his arm makes an angle of 15 degree with respect to the horizontal line. We can still calculate the arm length, or the horizontal distance from the trophy to the champion:

$L_2 = Lcos(15^0) = 0.655cos(15^0) = 0.655*0.966 = 0.633 m$