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exis [7]
2 years ago
15

a sloution has a very high hydroxide ion concentration.based on this what would your predit about this sloution

Physics
1 answer:
Makovka662 [10]2 years ago
7 0

Answer: t has a very low pH. It will have a very slippery feeling.

Explanation: It has high concentration of hydroxide ions · Explanation: pH is the negative logarithm of hydrogen in concentration

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A hockey puck is struck so that it slides at a constant speed and strikes the far side of the rink, 58.2 m away. The shooter hea
DENIUS [597]

Answer:

v = 33.66 m/s

Explanation:

Let hockey puck is moving at constant speed v

so here we have

d = vt

so time taken by the puck to strike the wall is given as

t = \frac{58.2}{v}

now time taken by sound to come back at the position of shooter is given as

t_2 = \frac{58.2}{340}

t_2 = 0.17s

so we know that total time is 1.9 s

1.9 = t + t_2

1.9 = t + 0.17

1.9 - 0.17 = t

t = 1.73 s

now we have

1.73 = \frac{58.2}{v}

v = 33.66 m/s

7 0
3 years ago
The balance Lenght of a potent ometer wire for a Cell of emf 1.62v is 90cm. if the Cell is replaced by another one of emf 1.08v.
andrew-mc [135]

Answer:

answer is 3.05v

Explanation:

hope it is helpful and briliant

6 0
2 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the
Rufina [12.5K]

Answer:

The  angular speed is w = 5.89 \ rad/s

Explanation:

From the question we are told that

    The time taken is  t = 1.6 s

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         \theta  =  n *  2 *  \pi

substituting values

        \theta  =  1.5 *  2 *  3.142

       \theta  = 9.426 \ rad

 The angular speed is mathematically represented as

         w =  \frac{\theta }{t}

substituting values

         w =  \frac{9.426}{1.6}

          w = 5.89 \ rad/s

     

3 0
3 years ago
Assume that you can drive at a constant speed of 100 kilometers per hour. suppose you started driving from the sun. how long wou
belka [17]
The Sun is 149.6 million kilometers from the earth.
There are 8760 hours in a year. 
876000 km are traveled in a year
It would take 170.776 years to reach the sun, or 171 years rather
4 0
2 years ago
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