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Mashutka [201]
3 years ago
10

Define capillarity ??​

Physics
2 answers:
earnstyle [38]3 years ago
8 0

Answer:

The tendency of a liquid in a capillary tube or absorbent material to rise or fall as a result of surface tension.

Anastasy [175]3 years ago
7 0

Answer:tendency of a liquid in a capillary tube

Explanation:

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A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy
GREYUIT [131]

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

8 0
3 years ago
A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric
sp2606 [1]

Answer:

a)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)E=0

c)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a)    r < a

At a radius r ,from gauss theorem

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)   a < r < b

E.ds=\dfrac{q_i}{\varepsilon _o}

The total induce in this surface = - Q+ Q =0

E.ds=\dfrac{0}{\varepsilon _o}

E = 0

c)   r > b

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

4 0
3 years ago
21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fric
jeyben [28]

Answer:

y = 12.82 m

Explanation:

We can solve this exercise using the energy work theorem

          W = ΔEm

friction force work is

          W = fr . s = fr s cos θ

the friction force opposes the movement, therefore the angle is 180º

           W = - fr s

we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular

           N -Wy = 0

           N = mg cos θ

the friction force remains

            fr = μ N

            fr = μ mg cos θ

             

work gives

           W = - μ mg s cos θ

initial energy

           Em₀ = ½ m v²

the final energy is zero, because it stops

we substitute

          - μ m g s cos θ = 0 - ½ m v²

          s = ½ v² / (μ g cos θ)

         

let's calculate

          s = ½ 20² / (0.55 9.8 cos 20)

          s = 39.49 m

this is the distance it travels along the plane, to find the vertical distance let's use trigonometry

            sin 20 = y / s

           y = s sin 20

           y = 37.49 sin 20

           y = 12.82 m

8 0
4 years ago
A car with a mass of 600 kg is traveling at a velocity of 10 m/s. How much kinetic energy does it have? The car has J of kinetic
konstantin123 [22]
The kinetic energy of an object is given by:
K= \frac{1}{2}mv^2
where m is the mass of the object and v its velocity. 

The car in this problem has a mass of m=600 kg and a velocity of v=10 m/s, therefore if we put these numbers into the equation, we find the kinetic energy of the car:
K= \frac{1}{2}mv^2= \frac{1}{2}(600 kg)(10 m/s)^2=30000 J
5 0
3 years ago
Sodium undergoes beta decay and becomes magnesium, write an equation. please help...
Tatiana [17]

Remark

There are 2 kinds of Beta Decay. This one is an example of a neutron becomes a proton and the emitted  particle is an electron.

_{11}^{23}\text{Na}\longrightarrow_{12}^{23}\text{Mg}+_{-1}^{ 0}\text{1e-}



6 0
3 years ago
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