1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Maslowich
3 years ago
12

An aluminum sphere (specific gravity = 2.70) falling through water reaches a terminal speed of 4.58 cm/s. What is the terminal s

peed of an air bubble of the same radius rising through water? Assume viscous drag in both cases and ignore the possibility of changes in size or shape of the air bubble; Density of air at 20°c is 1.20 kg/m3 and that of water is 998.21 kg/m3.
Physics
1 answer:
enyata [817]3 years ago
3 0

Answer:2.69 cm/s

Explanation:

We know that drag force

Fd=6 πμ r v

μ=Dynamic viscosity

r=radius

v=terminal velocity

Bouncy force

Fb= ρ V g

rho_w=density\ of\ water

V=volume

At the condition of terminal velocity

when going down ( aluminum)

Fb+ Fd= m g

 Fd_1 = m g - Fb_1---------1

when going up ( air)

Fb_2= mg +Fd

Fd_2= Fb_2 - mg-----------2

m =mass of object

m= density\times volume

\rho_a=density\ of\ Aluminium

\rho _b=density\ of\ bubble

Divide 1 and 2

\dfrac{v_1}{v_2}=\dfrac{mg-F_b_1}{F_b_2-mg}

\dfrac{4.58}{v_2}=\dfrac{\rho _a-\rho _w}{\rho _w-\rho _b}

divide by  \rho _w in R.H.S

\dfrac{4.58}{v_2}=\dfrac{2.7-1}{1.-0.0012}

v_2=\dfrac{4.58}{1.7}=2.69\ cm/s

You might be interested in
A car travels at a speed of 40 m/s for 29.0 s;<br>what is the distance traveled by the car?
lianna [129]

Answer: 1160 m

Explanation:

Speed = distance / time. Plug in 40 m/s for speed and 29 s for time in order to get the distance, 1160 m.

8 0
2 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
Which THREE forms of light are invisible light?
PtichkaEL [24]

Answer:

infrared light

Explanation:

3 0
3 years ago
Read 2 more answers
Where does energy come from that fuels our physical activity
joja [24]

Energy comes from the Sun, plants and animals that we eat, to provide us with nutrients we need for energy.


7 0
3 years ago
Read 2 more answers
A crate is pushed horizontally by a horizontal force 527.018 N . Sliding friction resists the motion, and the kinetic coefficien
gulaghasi [49]

Answer:

m= 10 kg a = 52 m / s²

Explanation:

For this problem we must use Newton's second law, let's apply it to each axis

X axis

      F - fr = ma

The equation for the force of friction is

    -fr = miu N

Axis y

     N- W = 0

     N = mg

Let's replace and calculate laceration

     F - miu (mg) = ma

    a = F / m - mi g

    a = 527.018 / m - 0.17 9.8

We must know the mass of the body suppose m = 10 kg

    a = 527.018 / 10 - 1,666

    a = 52 m / s²

5 0
4 years ago
Other questions:
  • Identify the false statement: Select one:
    12·2 answers
  • An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.5 kg mass of the pulley is concentrated on i
    14·1 answer
  • Why does the momentum before a collision equal the momentum after a collision?
    9·2 answers
  • What feature of a chemical equation represents the law of conservation of matter?
    12·2 answers
  • How much kinetic energy does a baseball with a mass of 0.143 kg have it it is traveling at a velocity of 41.1 m/s?
    8·1 answer
  • A roller coaster starts from rest at point a and continues to point b on a frictionless track. what best describes the changes i
    9·2 answers
  • Solve this question with explanation <br> Best answer will be brainliest <br> 40 points given
    7·2 answers
  • Give two examples of workplace environments where considerations must be made with respect to the possibility of electric discha
    9·1 answer
  • What is the mass of 2.5 mol of Ca, which has a molar mass of 40 g/mol?
    11·1 answer
  • Define a dipole. Hence, write the expression for calculating the electric moment of a dipole​
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!