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77julia77 [94]
3 years ago
9

Consider three drinking glasses. All three have thesame area base, and all three are filled to the same depth withwater. Glass A

is cylindrical. Glass B is wider at the top than atthe bottom, and so holds more water than A. Glass C is narrower atthe top than at the bottom, and so holds less water than A. Whichglass has the greatest liquid pressure at the bottom?
a. Glass A
b. Glass B
c. Glass C
d. All three have equal non-zero pressure at the bottom.
e. All three have zero pressure at the bottom.
Physics
1 answer:
Sergio [31]3 years ago
8 0

Answer:

The correct answer is C.  All three have equal non-zero pressure

Explanation:

Pressure is the relationship between the force and the area of ​​a body, when the bodies are liquid the formula that

         P = rho g h

Where rho is the density and h the height of the liquid

We see that for this expression the pressure does not depend on the shape of the container, but on its height, as the three vessels have the same height, the pressure at the bottom is the same.

The correct answer is C   All three have equal non-zero pressure

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Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height h ab
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Answer:

EXplained

Explanation:

from conservation of energy

change in potential energy = gain in kinetic energy

so as all he balls are throws from the same height thus the change in potential energy is the same for all the balls thus the gain in kinetic energy is the same for all the balls and as they have the same initial velocity thus the final velocity is the same for all the balls.

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While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far
viktelen [127]

Answer:

1.7 m

Explanation:

v_x = Velocity of ball in x direction = 4.47 m/s

u_y = Velocity of ball in y direction = 0

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

t = Time taken

s_y = Vertical displacement = 0.7 m

s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}

Horizontal displacement is given by

s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}

The passenger should throw the ball 1.7 m in front of the bucket.

5 0
3 years ago
A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
After today i might not be here no more
Norma-Jean [14]

Answer:

aw why? are you deleting the app for school?

6 0
2 years ago
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