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2 years ago

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An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, fil

led with a less dense liquid. What would you observe?

1 answer:

Anestetic [448]2 years ago

4 0

Answer:

The fraction of its volume inside liquid is increased .

Explanation:

According to principle pf floatation , an object floats on the surface of water

when the weight of liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid .

In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid is increased .

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Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

2 years ago

If a 1 kg book has 46 Joules of gravitational potential energy how high is the shelf it is on?

Mashcka [7]

Answer:

4.7m

Explanation:

Given parameters:

Mass of the book = 1kg

Gravitational potential energy = 46J

Unknown:

Height of the shelf = ?

Solution:

The potential energy is due to the position of a body above the ground.

Gravitational potential energy = mgh

m is the mass,

g is the acceleration due gravity = 9.8m/s²

h is the height which is unknown

46 = 1 x 9.8 x h

h = 4.7m

4 0

2 years ago

Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a

Yuliya22 [10]

Answer:

The value is $q = 3.4 *10^{-6} \ C$

Explanation:

From the question we are told that

The mass of each sphere is $m_1 = m_2 = m = 0.020 \ kg$

The length of the string is $l = 0.18 \ m$

The angle of with the vertical is $\theta = 7^o$

The acceleration due to gravity is $g = 9.8 \ m/s^2$

Generally the force acting between the forces is mathematically represented as

$F = T cos \theta = \frac{k* q^2}{ r^2}$

=> $T cos \theta = \frac{k* q^2}{ r^2}$

Generally from Pythagoras theorem the radius of the circular curve created by the force is

$r = 2 L sin (\theta )$

=> $r = 2* 0.180 sin (7)$

=> $r = 0.043 \ m$

=> $q = tan \theta * \frac{m * g * r^2 }{k}$

=> $q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}$

=> $q = 3.4 *10^{-6} \ C$

7 0

2 years ago

When water freezes, its volume increases by 9.05%. What force per unit area is water capable of exerting on a container when it

AnnZ [28]

Answer:

The Pressure is 0.20 MPa.

(b) is correct option.

Explanation:

Given that,

Change in volume = 9.05%

{tex]\dfrac{\Delta V}{V_{0}}=0.0905[/tex]

We know that.

The bulk modulus for water

$B=0.20\times10^{10}\ N/m^2$

We need to calculate the pressure difference

Using formula bulk modulus formula

$B=\Delta P\dfrac{V_{0}}{\Delta V}$

$\Delta P=B\dfrac{\Delta V}{V_{0}}$

$\Delta P=0.2\times10^{10}\times0.0905$

$\Delta P=0.2\times10^{6}\ Pa$

$\Delta P=0.20 MPa$

Hence, The Pressure is 0.20 MPa.

6 0

3 years ago

I wonder if people that say that there ugly is to get attention (not to be mean...)

alexandr1967 [171]

They do. mostly because they want validation from others

5 0

2 years ago

Read 2 more answers