You throw a ball vertically upward, and as it leaves your hand, its speed is 10.0 m/s.

An apple is held completely submerged just below the surface of water in a container. The apple is then moved to a deeper point

Tresset [83]

Answer:

Explanation:

When the apple is held submerged in water , it experiences a buoyant force due to which it floats in water . One has to apply downward force to keep it submerged. The lower the buoyant force , lower the force needed to submerge it in water.

When apple is held at much deeper point , it experience greater pressure due to column of water around it . So its size or its volume decreases . But its weight remains the same . Due to less volume , buoyant force also decreases ( buoyant force is equal to weight of displaced volume of water. )

Due to buoyant force becoming less , force needed on apple in downward direction will also be less.

4 0

3 years ago

two billiard balls moving along the same line hit each other head-on. each has a mass of 0.220 kg; one has an initial velocity o

Tems11 [23]

Hi there!

Since the collision is elastic, we must also satisfy the following condition:

Ei = Ef, or:

KEi = KEf

Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.

mv1 + mv2 = mvf1 + mvf2

0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)

0.2882/0.220 = vf1 + vf2

1.31 = vf1 + vf2

Now, we can express this as a conservation of energy:

1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²

Plug in values and simplify:

0.403315 = 1/2m(vf1² + vf2²)

Simplify further:

3.6665 = vf1² + vf2²

Use the equation derived from momentum above and solve for one variable:

vf2 = 1.31 - vf1

Plug in this expression for vf2:

3.6665 = vf1² + (1.31 - vf1)²

Expand:

3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²

Simplify:

1.9504 = -2.62vf1 + 2vf1²

Solve for vf1 using a graphing calculator:

vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.

Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.

Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):

vf2 = 1.31 - (-0.53) = 1.84 m/s.

Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.

7 0

3 years ago

Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10

USPshnik [31]

Answer:

D.

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

$y = \frac{1}{2} * g * t^{2}$

where y = height = 40 ft.

As all the parameters are given in SI units, it is advisable to convert this value to m, as follows:

$y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m$

Now, we can solve for t, as follows:

$t = \sqrt{\frac{2*y}{g}} = \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s$

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

$y = v0*t +\frac{1}{2} * g * t^{2}$

This leaves us a quadratic equation on t, as follows:

$t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s$

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s

5 0

3 years ago

QuestionDetails:

sleet_krkn [62]

To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

$r = \sqrt{x^2+R^2}$

Then the potential energy is

$U = \frac{-GMm}{\sqrt{x^2+R^2}}$

PART A) The force is excepted to be along x-axis.

Therefore we take a derivative of U with respect to x.

$F = -\frac{dU}{dx}$

$F = -\frac{d}{dx}(GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}))$

$F = \frac{GMmx}{(x^2+R^2)^{3/2}}$

This expression is the resultant magnitude of the Force F.

PART B) The magnitude of loss in potential energy as the particle falls to the center

$U = GMm(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})$

According to conservation of energy,

$\frac{1}{2}mv^2 = GMm (\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}})$

$\therefore v = \sqrt{2GM(\frac{1}{R}-\frac{1}{x^2+R^2})}$

7 0

3 years ago

What term is used to describe how well liquids flow

iris [78.8K]

Well water or any liquid actually takes the form of the cup of bowl or whatever you put the liquid in
REMEMBER- Liquid has no shape.
hope this helps!!!!!!!!

8 0

3 years ago

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