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Oksana_A [137]
3 years ago
15

A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it

hits the water?is it 14 m/s, 9.9 m/s, or 4.8 m/s
Physics
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

14 m/s

Explanation:

The problem can be solved by using the law of conservation of energy. In fact:

- The mechanical energy of the stone at the top of the cliff is just gravitational potential energy (because the stone is at rest, so its kinetic energy is zero), and it is given by

E=U=mgh (1)

where m is the mass of the stone, g=9.8 m/s^2 is the acceleration due to gravity, and h=10 m is the height of the cliff.

- The mechanical energy of the stone when it hits the water is just kinetic energy (because the height of the stone has become zero, so the gravitational potential energy is zero), so it is

E=K=\frac{1}{2}mv^2 (2)

where v is the speed of the stone when it hits the water.

Since the mechanical energy is conserved, we can equalize (1) and (2), and solving for v we find:

mgh=\frac{1}{2}mv^2\\v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(10 m)}=14 m/s

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1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

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A visitor to the observation deck of a skyscraper manages to drop a penny over the edge. As the penny falls faster, the force du
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If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
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A larger object is made of two balls separated by a massless rigid rod that is 1.5 m long. The mass of the red ball at one end i
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Answer:

The speed of the 1 kg red ball 8.04 m/s .

Explanation:

Given :

Separation between rods , d = 1.5 m .

Mass of the red ball is 1 kg .

Mass of the orange ball is 5.7 kg .

Angular velocity , \omega=1\ rev/s = 2\pi\ rad/s .

Now , distance of center of mass  from red ball is :

r =\dfrac{0\times 1+1.5\times 5.7}{1+5.7}\\\\r = 1.28\ m

We know , speed is given by :

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A person with a mass of 72 kg is riding a bicycle is accelerating at a rate of 5m/s on a horizontal surface. What is the weight
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Answer:

w = 706.32 [N]

Explanation:

The force due to gravitational acceleration can be calculated by means of the product of mass by gravitational acceleration.

w = m*g

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g = gravity acceleration = 9.81 [m/s²]

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w = 72*9.81\\w = 706.32 [N]

7 0
3 years ago
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